16: Acids and Bases

Downloadable files

• Please print and work out the question before looking at the answers
• Chapter 16: Acids and Bases pdf

16.1: Brønsted-Lowry Concept of Acids and Bases

Textbook: Section 16.1

Arrhenius & Brønsted Acids/Bases

Exercise \(\PageIndex{1.a}\)

In the reaction, identify the acid and base.

\(NH_{3}(g)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq)+OH^{-}(aq)\)

Answer

In this reaction H₂O is donating a proton so it is the acid

\(NH_{3}\) is accepting the proton so it is the base

Exercise \(\PageIndex{1.b}\)

What is the conjugate acid of \(NH_{3}\)?

Answer

A conjugate acid is a compound formed when an acid donates a proton to a base. So basically it is a base with a hydrogen ion added to it

\(NH_{3}(g)+H^{+}\rightleftharpoons NH_{4}^{+}(aq)\nonumber \)

So the conjugate acid of \(NH_{3}\) is \(NH_{4}^{+}\)

Exercise \(\PageIndex{1.c}\)

What is the conjugate base of H₂O?

Answer

A conjugate base is what is left over after an acid has donated a proton. 

\(H_{2}O\rightleftharpoons H^{+} + OH^{-} \nonumber\)

So the Conjugate base of H₂O is \(OH^{-}\)

Exercise \(\PageIndex{1.d}\)

What is the conjugate acid of NaHSO₃?

Answer

NaHSO₃ is an amphoteric substance, which means it can be an acid or a base

\(HSO_{3}^{-}+H^{+}\rightarrow H_{2}SO_{3} \)

So the conjugate acid of \(NaHSO_{3}\) is \(H_{2}SO_{3}\)

Exercise \(\PageIndex{1.e}\)

What is the conjugate base of NaHSO₃?

Answer

NaHSO₃ is an amphoteric substance, which means it can be an acid or a base

\(HSO_{3}^{-}+OH^{-}\rightarrow SO_{3}^{-2}+H_{2}O\nonumber\)

So the conjugate base of \(NaHSO_{3}\) is \(SO_{3}^{-2}\)

 

Exercise \(\PageIndex{1.f}\)

It is known that the hydride ion H^- is a stronger base than OH^-, what is(are) the product(s) of the reaction: H^-(aq) +H₂O(l) 

Answer

\(H^{-}(aq)+H_{2}O(l)\rightarrow H_{2}(g)+OH^{-}(aq)\nonumber\)

base        acid        C.A.        C.B.

Since H^- is a stronger base than OH^-, the reaction is reasonable to take place.

 

16.2: Water and the pH Scale

Textbook: Section 16.2

pH and Strong Acids/Bases

Exercise \(\PageIndex{2.a}\)

In a sample of lemon juice, [H+] is 6.2x10^-4 M.  What is the pH?

Answer

\[pH=-log[H^{+}]=-log[6.2*10^{-4}M]=3.2\nonumber \]

Exercise \(\PageIndex{2.b}\)

A sample of detergent has a pH of 8.20.  What is the [H⁺]?

Answer

\[[H^{+}]=10^{-pH}=10^{-8.20}=6.3*10^-9M\nonumber \]

Exercise \(\PageIndex{2.c}\)

What is [OH^-] of the detergent in Q 16.2.b?

Answer

\[[OH^{-}]=\frac{K_{w}}{[H^{+}]}=\frac{10^{-14}}{6.3*10^{-9}}=1.6*10^{-6}\nonumber \]

Exercise \(\PageIndex{2.d}\)

What is the pH of 0.055M of HCl?

Answer

\[H^{+}=0.055M\nonumber\]

\[pH=-log[H^{+}]=-log[0.055M]=1.26\nonumber \]

Exercise \(\PageIndex{2.e}\)

What is the pH of 0.001M of Ba(OH)₂?

Answer

\[[OH^{-}]=0.001*2=0.002M\nonumber \]

\[[H^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{10^{-14}}{2.0*10^{-3}}=5.0*10^{-12}M\nonumber \]

\[pH=-log[H^{+}]=-log[5.0*10^{-12}M]=11.30\nonumber \]

Exercise \(\PageIndex{2.f}\)

What is the pOH of the solution in Q 16.2.e?

Answer

\[[OH^{-}]=0.001*2=0.002M\nonumber \]

\[pOH=-log[OH^{-}]=-log[2.0*10^{-3}M]=2.70\nonumber \]

Exercise \(\PageIndex{2.g}\)

If 0.56g of CaO is dissolved in the water to make 1.0L solution.  What is the pH of the solution?

Answer

\[CaO(s)+H_{2}O(l)\rightleftharpoons Ca(OH)_{2}(aq)\nonumber \]

\[\frac{\frac{0.56}{56\,g/mol}*\frac{1\,mol\,Ca(OH)_{2}}{1\,mol\,CaO}}{1.0L}=0.01\,M\,Ca(OH)_{2}\nonumber \]

\[[OH^{-}]=0.01*2=0.02M\nonumber \]

\[pOH=-log[OH^{-}]=-log[2.0*10^{-2}M]=1.70\nonumber \]

\[pH=14-pOH=14-1.70=12.3\nonumber \]

 

16.3: Equilibrium Constants for Acids and Bases

Textbook: Section 16.3

Weak Acids

Exercise \(\PageIndex{3.a}\)

What is the pH of 0.20M aqueous HF?  K_a = 6.8x10^-4

Answer

\[HF(aq)\rightleftharpoons H^{+}(aq)+F^{-}(aq) \nonumber \]

\[[HF]_{i}>100K_{a} \nonumber \]

Therefore,

\[[H^{+}]=\sqrt{K_{a}[HF]_{i}}=\sqrt{(6.8*10^{-4})*0.20}=1.2*10^{-2}M \nonumber \]

\[pH=-log[H^{+}]=-log[1.2*10^{-2}M]=1.93 \nonumber \]

Exercise \(\PageIndex{3.b}\)

What is the pH of 0.0050M aqueous HF?  K_a = 6.8x10^-4

Answer

R	HF(aq)	⇌	H+(aq)          +	F-(aq)
I	0.0050		0	0
C	-x		+x	+x
E	0.0050-x		x	x

\(K_{a}=\frac{x^{2}}{0.0050-x}\) Note 100K_a is not less than [HA]_i so we need to use the quadratic formula.

\[K_{a}\left ( 0.0050-x \right )=x^{2}\Rightarrow x^{2}+K_{a}x-K_{a}\left ( 0.0050 \right ) \nonumber \]

\[x^{2}+0.00068x-3.4*10^{-5} \nonumber \]

\[x=\frac{-0.00068\pm \sqrt{\left ( 0.00068 \right )^{2}-4(1)(-3.4*10^{-6})}}{2} \nonumber \]

\[x=\frac{-0.00068\pm 0.00375}{2} \nonumber \]

\[x=[H^{+}]=0.001535M \nonumber \]

\[pH=-log[H^{+}]=-log[0.001535M]=2.8 \nonumber \]

Exercise \(\PageIndex{3.c}\)

Which one is more acidic, 0.2MHF or 0.02MHCl?

Answer

\[0.02HCl:\,pH=-log[0.02]=1.70 \nonumber \]

\[0.2HF:\,pH=-log(\sqrt{K_{a}[HA]_{i}})=-log(\sqrt{7.2*10^{-4}[0.2]})=1.93 \nonumber \]

Exercise \(\PageIndex{3.d}\)

What is the pH of 0.04M of NH₄Cl?  K_a=5.6x10^-10

Answer

\[NH_{4}^{+}(aq)\rightleftharpoons NH_{3}(aq)+H^{+}(aq)\,\,\,K_{a}=5.6*10^{-10} \nonumber \]

\[\left [NH_{4}^{+}\right ]_{i}>100K_{a}\nonumber\]

Therefore,

\[[H^{+}]=\sqrt{K_{a}[NH_{4}^{+}]_{i}}=\sqrt{5.6*10^{-10}*(0.04)}=4.73*10^{-6}M \nonumber \]

\[pH=-log[H^{+}]=-log[4.73*10^{-6}M]=5.32 \nonumber \]

Exercise \(\PageIndex{3.e}\)

A 0.10M solution of lactic acid (HC₃H₅O₃, one acidic hydrogen) has a pH of 2.45.  What is the K_a for lactic acid?

Answer

\[\left [ H^{+} \right ]=10^{-2.45}=0.0035M \nonumber \]

R	HC3H5O3(aq)	⇌	C3H5O3^-(aq)          +	H^+(aq)
I	0.10M		0	0
C	-x		+x	+x
E	0.10-x		x	x

\[K_{a}=\frac{x^{2}}{0.10-x}=\frac{0.0035^{2}}{0.10-0.0035}=1.30*10^{-4} \nonumber \]

 

Weak Base

Exercise \(\PageIndex{3.f}\)

What is the concentration of OH^- in 0.10M of ethylamine (C₂H₅NH₂)? Kb=6.4x10^-4

Answer

\[\left [ C_{2}H_{5}NH_{2} \right ]>100K_{b} \nonumber \]

\[\left [ OH^{-} \right ]=\sqrt{\left (6.4*10^{-4}\right )*0.10}=0.008M \nonumber \]

Exercise \(\PageIndex{3.g}\)

What is the concentration of OH^- in 0.005M of ethylamine (C₂H₅NH₂)?

Answer

R	C2H5NH2(aq)     +	H2O(l)	⇌	C2H5NH3^+(aq)     +	OH^-(aq)
I	0.005M			0	0
C	-x			+x	+x
E	0.005-x			x	x

\[K_{b}=\frac{x^{2}}{0.005-x}=6.4*10^{-4} \nonumber \]

\[x=1.5*10^{-3} \nonumber \]

Exercise \(\PageIndex{3.h}\)

Which solution is more basic, 0.10M of ethylamine or 0.01M of NaOH?

Answer

NaOH, 0.01M of NaOH has a pOH of 2.00

Exercise \(\PageIndex{3.i}\)

A solution of NH₃ has a pH of 10.25.  What is the concentration of the solution? Kb=1.8x10^-5

Answer

\[10^{-10.25}=5.6*10^{-11} \nonumber \]

R	NH3(aq)     +	H2O(l)	⇌	NH4^+(aq)	OH^-(aq)
I	Unknown			0	0
C	x			+x	+x
E	Unknown			x	x

\[x=\left [ OH^{-} \right ]=\frac{10^{-14}}{5.6*10^{-11}}=1.8*10^{-4} \nonumber \]

\[K_{b}=\frac{x^{2}}{Unknown-x}=1.8*10^{-5}=\frac{\left (1.8*10^{-4}\right )^{2}}{U-\left (1.8*10^{-4}\right )} \nonumber \]

\[U=2.0*10^{-3}M \nonumber \]

Exercise \(\PageIndex{3.j}\)

If 1.06g of Na₂CO₃ is dissolved in plenty of water to make 1.0L of the solution, what is the pH of the solution? Kb=1.8x10^-4

Answer

\[\frac{\frac{1.06g}{106g/mol}}{1.0L}=0.01M \nonumber \]

R	CO3^2-(aq)     +	H2O(l)	⇌	HCO3^-(aq)     +	OH^-(aq)
I	0.01M			0	0
C	-x			+x	+x
E	0.01-x			x	x

\[K_{b}=\frac{x^{2}}{0.01-x}=1.8*10^{-4} \nonumber \]

\[x=1.3*10^{-3}M \nonumber \]

\[pOH=2.90 \nonumber \]

\[pH=14.00-2.90=11.10 \nonumber \]

 

Polyprotic Acids

Exercise \(\PageIndex{3.k}\)

What is the pH of a 0.002M solution of H₂CO₃? Ka₁=4.3x10^-7, Ka₂ = 5.6x10^-11

Answer

\[Ka_{1}=4.3*10^{-7},\,Ka_{2}=5.6*10^{-11} \nonumber \]

\[\frac{Ka_{1}}{Ka_{2}}>1000 \nonumber \]

\[\left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=\sqrt{Ka_{1}*0.002}=2.9*10^{-5}M \nonumber \]

\[pH=4.53 \nonumber \]

Exercise \(\PageIndex{3.l}\)

What is the concentration of CO₃^2- ion in the solution in Q16.3.11?

Answer

\[\left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=2.9*10^{-5}M \nonumber \]

R	HCO3^-(aq)	⇌	H^+(aq)     +	CO3^2^-(aq)
I	2.9*10^-5		2.9*10^-5	0
C	-x		-x	+x
E	2.9*10^-5-x		2.9*10^-5-x	x

\[Ka_{2}=\frac{\left (2.9*10^{-5}-x\right )x}{2.9*10^{-5}-x}=5.6*10^{-11} \nonumber \]

\[\left [CO_{3}^{2-}\right ]=5.6*10^{-11}M \nonumber \]

Exercise \(\PageIndex{3.m}\)

What is the pH of a 0.05M of sulfurous acid (H₂SO₃)? Ka₁=1.7x10^-2, Ka₂=6.4x10^-8

Answer

\[\frac{Ka_{1}}{Ka_{2}}>1000 \nonumber \]

R	H2SO3(aq)	⇌	H^+(aq)	HSO3^-(aq)
I	0.05		0	0
C	-x		+x	+x
E	0.05-x		x	x

\[Ka_{1}=\frac{x^{2}}{0.05-x}=1.7*10^{-2} \nonumber \]

\[x=2.2*10^{-2}M \nonumber \]

\[pH=1.66 \nonumber \]

Exercise \(\PageIndex{3.n}\)

What is the concentration of SO₃^2- ion in the solution in Q16.3.13?

Answer

\[\left [H^{+}\right ]=\left [HSO_{3}^{-}\right ]=2.2*10^{-2}M \nonumber \]

\[Ka_{2}=\frac{\left (2.2*10^{-2}-x\right )x}{2.2*10^{-2}-x}=6.4*10^{-8} \nonumber \]

\[\left [SO_{3}^{2-}\right ]=6.4*10^{-8}M \nonumber \]

Exercise \(\PageIndex{3.o}\)

What is the concentration of a sample solution of H₂CO₃ that has a pH = 4.50?

Answer

\[10^{-4.50}=3.16*10^{-5} \nonumber \]

\[\left [H^{+}\right ]=\sqrt{Ka_{1}*\left [H_{2}CO_{3}\right ]}=3.16*10^{-5}M \nonumber \]

\[Ka_{1}=4.3*10^{-7} \nonumber \]

\[\left [H_{2}CO_{3}\right ]=0.0023M \nonumber \]

 

16.4: Acid-Base Properties of Salts

Textbook: Section 16.4

Percent Ionization

Exercise \(\PageIndex{4.a}\)

What is the percent ionization of 0.25 aqueous HF?  Ka = 6.8x10^-4

Answer

\[HF(aq)\rightleftharpoons H^{+}(aq)+F^{-}(aq)\nonumber \]

\[\left [HF\right ]_{i}>100Ka\nonumber\]

Therefore,

\[\left [H^{+}\right ]=\sqrt{Ka\left [HF\right ]_{i}}=\sqrt{6.8*10^{-4}*0.25}=1.3*10^{-2}M\nonumber\]

\[\frac{1.3*10^{-2}}{0.25}=5.2\%\nonumber\]

Exercise \(\PageIndex{4.b}\)

What is the percent ionization of 0.0055 aqueous HF?  Ka = 6.8x10^-4

Answer

R	HF(aq)	⇌	H^+(aq)     +	F^-(aq)
I	0.0055M		0	0
C	-x		+x	+x
E	0.0055-x		x	x

\[Ka=\frac{x^{2}}{0.0055-x}=6.8*10^{-4}\nonumber\]

\[x=\left [H^{+}\right ]=1.60*10^{-3}M\nonumber\]

\[\frac{1.60*10^{-3}}{0.0055}=29.5\%\nonumber\]

Exercise \(\PageIndex{4.c}\)

What is the percent ionization of 0.05M of NH₄Cl?  Ka=5.6x10^-10

Answer

\(NH_{4}^{+}(aq)\rightleftharpoons NH_{3}(aq)+H^{+}(aq)\), \(Ka=5.6*10^{-10}\)

\[\left [NH_{4}^{+}\right ]_{i}>100Ka\nonumber\]

\[\left [H^{+}\right ]=\sqrt{Ka\left [NH_{4}^{+}\right ]_{i}}=\sqrt{5.6*10^{-11}*0.005}=5.29*10^{-6}M\nonumber\]

\[\frac{5.29*10^{-6}}{0.05}*100\%=0.011\%\nonumber\]

Exercise \(\PageIndex{4.d}\)

What is the percent ionization of a 0.002M solution of H₂CO₃? Ka₁=4.3x10^-7, Ka₂ = 5.6x10^-11

Answer

\[Ka_{1}=4.3*10^{-7}, Ka_{2}=5.6*10^{-11}\nonumber\]

\[\frac{Ka_{1}}{Ka_{2}}>1000 \left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=\sqrt{Ka_{1}*0.002}=2.9*10^{-5}M\nonumber\]

\[\frac{2.9*10^{-5}}{0.002}*100\%=1.45\%\nonumber\]

Exercise \(\PageIndex{4.e}\)

What is the percent ionization of 0.04M of hydrazoic acid (HN₃)?  Ka=1.9x10^-5

Answer

\[HN_{3}(aq)\rightleftharpoons H^{+}(aq)+N_{3}^{-}(aq)\nonumber\]

\[\left [HN_{3}\right ]_{i}>100Ka\nonumber\]

Therefore,

\[\left [H^{+}\right ]=\sqrt{Ka\left [HN_{3}\right ]_{i}}=\sqrt{1.9*10^{-5}*0.04}=8.7*10^{-4}M\]

\[\frac{8.7*10^{-4}}{0.04}=2.2\%\nonumber\]

 

pH of Various Salts

Exercise \(\PageIndex{4.f}\)

What is the pH of 0.05M of NH₄Cl?  Ka=5.6x10^-10

Answer

\[NH_{4}^{+}(aq)\rightleftharpoons NH_{3}(aq)+H^{+}(aq)\), \(Ka=5.6*10^{-10}\nonumber \]

\[\left [NH_{4}^{+}\right ]_{i}>100Ka \nonumber \]

Therefore,

\[\left [H^{+}\right ]=\sqrt{Ka\left [NH_{4}^{+}\right ]_{i}}=\sqrt{5.6*10^{-10}*0.05}=5.3*10^{-6}M\nonumber \]

\[pH=-log\left (5.3*10^{-6}M\right )=5.28\nonumber \]

Exercise \(\PageIndex{4.g}\)

What is the pH of 0.05M of NaClO? Kb=3.3x10^-7

Answer

\[ClO^{-}(aq)+H_{2}O(l)\rightleftharpoons HClO(aq)+OH^{-}(aq)\),\(K_{b}=3.3*10^{-7}\nonumber \]

\[\left [ClO^{-}\right ]_{i}>100K_{b}\nonumber \]

\[\left [OH^{-}\right ]=\sqrt{K_{b}\left [ClO^{-}\right ]_{i}}=\sqrt{3.3*10^{-7}*0.05}=1.3*10^{-4}M\nonumber \]

\[pOH=-log\left (1.3*10^{-4}M\right )=3.89\nonumber \]

\[pH=14.00-3.89=10.11\nonumber \]

Exercise \(\PageIndex{4.h}\)

What is the pH of 0.05M of NaHCO₃? Kb=2.3x10^-8

Answer

\[HCO_{3}^{-}(aq)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq)+OH^{-}(aq)\), \(K_{b}=2.3*10^{-8}\nonumber \]

\[\left [HCO_{3}^{-}\right ]_{i}>100K_{b}\nonumber \]

Therefore,

\[\left [OH^{-}\right ]=\sqrt{K_{b}\left [HCO_{3}^{-}\right ]_{i}}=\sqrt{2.3*10^{-8}*0.05}=3.4*10^{-5}M\nonumber \]

\[pOH=-log\left (3.4*10^{-5}M\right )=4.47\nonumber \]

\[pH=14.00-4.47=9.53\nonumber \]

Exercise \(\PageIndex{4.i}\)

What is the pH of 0.05M of KF? Kb=1.5x10^-11

Answer

\[F^{-}(aq)+H_{2}O(l)\rightleftharpoons HF(aq)+OH^{-}(aq)\), \(K_{b}=1.5*10^{-11}\nonumber \]

\[\left [F^{-}\right ]_{i}>100K_{b}\nonumber \]

Therefore,

\[\left [OH^{-}\right ]=\sqrt{K_{b}\left [F^{-}\right ]_{i}}=\sqrt{1.5*10^{-11}*0.05}=8.7*10^{-7}M\nonumber \]

\[pOH=-log\left (8.7*10^{-7}M\right )=6.06\nonumber \]

\[pH=14.00-6.06=7.94\nonumber \]

Exercise \(\PageIndex{4.j}\)

How many grams of NaHCO₃ will be used to make a 1.0L solution that has a pH = 9.0?

Answer

\[\left [OH^{-}\right ]=10^{-\left (14.0-9.0\right )}=10^{-5}M\nonumber \]

R	HCO3^-(aq)     +	H2O(l)	⇌	H2CO3(aq)     +	OH^-(aq)
I	Unknown			0	0
C	-x			+x	+x
E	Unknown-x			x	x

\[K_{b}=\frac{x^{2}}{U-x}=\frac{\left (10^{-5}\right )}{\left (U-10^{-5}\right )}=2.3*10^{-8}\nonumber \]

\[U=\left [HCO_{3}^{-}\right ]=4.4*10^{-3}M\nonumber \]

\[4.4*10^{-3}M*1.0L=4.4*10^{-3}mol\nonumber \]

\[84.0g/mol*4.4*10^{-3}mol=0.37g\nonumber \]

 

16.5: Acid-Base Equilibrium Calculations

Textbook: Section 16.5

Acid Anhydrides

Exercise \(\PageIndex{5.a}\)

What is the pH of the solution if 0.05mol of SO₃ is dissolved in the water to make 1.0L solution?  (SO₃ is very soluble.)

Answer

\[SO_{3}(g)+H_{2}O(l)\rightleftharpoons H_{2}SO_{4}(aq) \nonumber \]

\[H_{2}SO_{4}(aq)\rightleftharpoons H^{+}(aq)+HSO_{4}^{-}(aq) \nonumber \]

\[\frac{0.05mol}{1.0L}=0.05M  pH=-log\left (0.05M\right )=1.30 \nonumber \]

Exercise \(\PageIndex{5.b}\)

How many grams of SO₃ is needed to make a 1.0L solution that has a pH=1.0?

Answer

\[SO_{3}(g)+H_{2}O(l)\rightleftharpoons H_{2}SO_{4}(aq) \nonumber \]

\[H_{2}SO_{4}(aq)\rightleftharpoons H^{+}(aq)+HSO_{4}^{-}(aq) \nonumber \]

\[\left [H^{+}\right ]=0.1M \nonumber \]

\[0.1M*1.0L=0.1mol \nonumber \]

\[0.1mol*80g/mol=8.0g \nonumber \]

Exercise \(\PageIndex{5.c}\)

What is the pH of the solution if 0.002mol of CO₂ is dissolved in the water to make 1.0L solution at 25^oC and 0.1atm?  (The solubility of CO₂ in pure water at 25^oC and 0.1atm is 0.0037M.) Ka₁=4.3x10^-7, Ka₂ = 5.6x10^-11

Answer

\[CO_{2}(g)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq) \nonumber \]

\[H_{2}CO_{3}(aq)\rightleftharpoons H^{+}(aq)+HCO_{3}^{-}(aq) \nonumber \]

\[\frac{0.002mol}{1.0L}=0.002M \nonumber \]

\[\left [H^{+}\right ]=\sqrt{4.3*10^{-7}*0.002}=2.93*10^{-5}M\]

\[pH=-log\left (2.93*10^{-5}M\right )=4.53 \nonumber \]

Exercise \(\PageIndex{5.d}\)

What is the concentration of CO₃^2- ion in the solution in Q 16.5.3?

Answer

\[\left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=2.93*10^{-5}M \nonumber \]

R	HCO3^-(aq)	⇌	H^+(aq)     +	CO3^2-(aq)
I	2.93*10^-5		2.93*10^-5	0
C	-x		+x	+x
E	2.93*10^-5-x		2.93*10^-5+x	x

\[Ka_{2}=\frac{\left (2.93*10^{-5}-x\right)x}{2.93*10^{-5}+x}=5.6*10^{-11} \nonumber \]

\[\left [CO_{3}^{2-}\right]=5.6*10^{-11} \nonumber \]

Exercise \(\PageIndex{5.e}\)

How many grams of CO₂ is needed at 25^oC and 0.1atm to make a 1.0L solution that has a pH=4.50?

Answer

\[pH=4.50 \nonumber \]

\[\left [H^{+}\right]=10^{-4.50}=3.16*10^{-5}M \nonumber \]

\[CO_{2}(g)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq) \nonumber \]

R	H2CO3(aq)	⇌	H^+(aq)     +	HCO3^-(aq)
I	M		0	0
C	-x		+x	+x
E	M-x		3.16*10^-5	x

\[Ka=\frac{\left ( 3.16*10^{-5} \right )^{2}}{M-3.16*10^{-5}}=4.3*10^{-7} \nonumber \]

\[M=2.35*10^{-3}mol/L \nonumber \]

\[2.35M*1.0L=2.36mol \nonumber \]

\[2.35mol*44g/mol=0.10g \nonumber \]

 

Basic Anhydrides

Exercise \(\PageIndex{5.f}\)

What is the pH of the solution that is prepared by dissolving 0.62g of Na₂O in enough water to make 1.0L?

Answer

\[Na_{2}O(s)+H_{2}O(l)\rightleftharpoons 2NaOH(aq) \nonumber \]

\[NaOH(aq)\rightleftharpoons Na^{+}(aq)+OH^{-}(aq) \nonumber \]

\[\left [NaOH\right ]=\left [OH^{-}\right ]=\frac{2*\frac{0.62g}{62g/mol}}{1.0L}=0.02M \nonumber \]

\[pH=14-log\left [OH^{-}\right ]=14.0-\left (-log\left (0.02M\right )\right )=12.3 \nonumber \]

Exercise \(\PageIndex{5.g}\)

How many grams of Na₂O is needed to make a 1.0L solution that has a pH=13.0?

Answer

\[pH=13.0 \nonumber \]

\[\left [OH^{-}\right]=10^{-\left (14-13\right )}=0.10M \nonumber \]

\[0.10M*1.0L=0.10mol \nonumber \]

\[Na_{2}O(s)+H_{2}O(l)\rightleftharpoons 2NaOH(aq) \nonumber \]

\[\frac{Na_{2}O}{2NaOH}=\frac{x mol}{0.10mol} \nonumber \]

\[x=0.05mol \nonumber \]

\[0.05mol*62g/mol=3.10g \nonumber \]

Exercise \(\PageIndex{5.h}\)

What is the pH of the solution that is prepared by dissolving 0.56g of CaO in the water to make 1.0L?

Answer

\[CaO(s)+H_{2}O(l)\rightleftharpoons Ca\left (OH\right )_{2}(aq) \nonumber \]

\[Ca\left (OH\right )_{2}(aq)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq) \nonumber \]

\[2\left [Ca\left (OH\right )_{2}\right]=\left [OH^{-}\right ]=\frac{2*\frac{0.56g}{56g/mol}}{1.0L}=0.02M \nonumber \]

\[pH=14-\left (-log\left [ OH^{-} \right ]\right )=14.0-1.70=12.3 \nonumber \]

Exercise \(\PageIndex{5.i}\)

How many grams of CaO is needed to make a 1.0L solution that has a pH = 13.0?

Answer

\[pH=13.0 \nonumber \]

\[\left [OH^{-}\right]=10^{-\left ( 14-13 \right )}=0.10M \nonumber \]

\[0.10M*1.0L=0.10mol \nonumber \]

\[CaO(s)+H_{2}O(l)\rightleftharpoons Ca\left (OH\right )_{2}(aq) \nonumber \]

\[Ca\left (OH\right )_{2}(aq)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq) \nonumber \]

\[\frac{2OH^{-}}{Ca\left (OH\right )_{2}}=\frac{Ca\left (OH\right )_{2}}{CaO}=\frac{0.10mol}{xmol} \nonumber \]

\[x=0.05mol \nonumber \]

\[0.05mol*56g/mol=2.80g \nonumber \]

Exercise \(\PageIndex{5.j}\)

If 100.0ml of the solution in Q 16.5.9 is transferred to a 500.0ml container, and plenty water was added to fill it up, what is the pH of the solution?

Answer

\[\left [OH^{-}\right]=10^{-\left ( 14-13 \right )}=0.10M \nonumber \]

\[Ca\left (OH\right )_{2}(aq)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq) \nonumber \]

\[\frac{\left [ Ca\left (OH\right )_{2} \right ]}{\left [ OH^{-} \right ]}=\frac{1}{2} \nonumber \]

\[\left [ Ca\left (OH\right )_{2} \right ]=0.05M \nonumber \]

\[\frac{0.05M*0.10L}{0.50L}=0.01M \left [ OH^{-} \right ]=0.02M \nonumber \]

\[pH=14-\left ( -log\left [ OH^{-} \right ] \right )=14.0-1.7=12.3 \nonumber \]

 

Polyprotic Acids

Exercise \(\PageIndex{5.k}\)

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the pH of H₂SeO₃?

Answer

\[pH=-log\left [ H^{+} \right ]=-log\left [ HSeO_{3}^{-} \right ]=-log\sqrt{2.7*10^{-5}\left ( 0.50M \right )}=2.43 \nonumber \]

Exercise \(\PageIndex{5.l}\)

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the concentration of H⁺?

Answer

\[pH=-log\left [ H^{+} \right ]=\left [ HSeO_{3}^{-} \right ]=\sqrt{2.7*10^{-5}\left ( 0.50M \right )}=0.0037M \nonumber \]

Exercise \(\PageIndex{5.m}\)

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the concentration of HSeO₃^-?

Answer

\[pH=-log\left [ H^{+} \right ]=\left [ HSeO_{3}^{-} \right ]=\sqrt{2.7*10^{-5}\left ( 0.50M \right )}=0.0037M \nonumber \]

Exercise \(\PageIndex{5.n}\)

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the concentration of H₂SeO₃?

Answer

\[H_{2}SeO_{3}=0.50M-\left [ H^{+} \right ]=0.50M-0.0037M=0.4963M=0.50M \nonumber \]

Exercise \(\PageIndex{5.o}\)

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the concentration of OH^-?

Answer

\[\left [ OH^{-} \right ]=\frac{K_{w}}{\left [ H^{+} \right ]}=\frac{10^{-14}}{\left [ H^{+} \right ]}=2.7*10^{-12}M \nonumber \]

Exercise \(\PageIndex{5.p}\)

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the concentration of SeO₃^2-?

Answer

\[Ka_{2}=\frac{\left [ H^{+} \right ]\left [ CO_{3}^{2-} \right ]}{\left [ HCO_{3}^{-} \right ]}=\left [ CO_{3}^{2-} \right ]=2.5*10^{-9}M \nonumber \]

 

16.7: Lewis Concept of Acids and Bases

Textbook: Section 16.7

Exercise \(\PageIndex{7.a}\)

CO can form complexes with metals, eg. Fe(CO)₅, Ni(CO)₄.  Is CO a lewis acid or a lewis base?

Answer

CO has a lone pair of electrons so it is a Lewis base

Exercise \(\PageIndex{7.b}\)

In Q 16.7.a, Is the metal, such as Fe(II), a lewis acid or a lewis base?

Answer

Fe(II) accepts the electron pair so it is a Lewis acid

Exercise \(\PageIndex{7.c}\)

In the reaction, \(Zn\left ( OH \right )_{2}(s)+2OH^{-}(aq)\rightleftharpoons Zn\left ( OH \right )_{4}^{2-}(aq)\), Which one is the lewis acid?

Answer

\(Zn\left ( OH \right )_{2}\) accepts an electron pair so it is a lewis acid

Exercise \(\PageIndex{7.d}\)

In the reaction, \(CO_{2}+O^{2-}\rightarrow CO_{3}^{2-}\), Which one is the lewis acid and lewis base?

Answer

Lewis acid: \(CO_{2}\)

Lewis base: \(O^{2-}\)

Exercise \(\PageIndex{7.e}\)

Can CH₃NH₂ be a lewis acid or a lewis base?

Answer

Lewis base

Exercise \(\PageIndex{7.f}\)

Trimethylamine (CH₃)₃N can react with diborane B₂H₆  after its dissociation to form (CH₃)₃N-BH₃.  Which one is the lewis acid?  Which one is the lewis base?

Answer

Lewis acid: B₂H₆

Lewis base: (CH₃)₃N

Exercise \(\PageIndex{7.g}\)

In the reaction, \(H_{2}NOH(aq)+HCl(aq)\rightarrow \left [ H_{3}NOH \right ]Cl(aq)\), which one is the lewis acid?

Answer

\(H_{2}NOH\)

Exercise \(\PageIndex{7.h}\)

In the reaction, \(SO_{2}(g)+BF_{3}(g)\rightarrow O_{2}S-BF_{3}(s)\), which one is the lewis acid? and the lewis base?

Answer

Lewis acid: BF₃

Lewis base: SO₂

 

General Questions

Textbook: Section 16

Exercise \(\PageIndex{a}\)

Which of the following would be considered a base according to the Brønsted-Lowry definition but not by the Arrhenius definition?

1. Ba(OH)₂ (aq)
2. HBr (g)
3. HF (g)
4. KOH (aq)
5. NH₃ (g)

Answer

e. NH₃ (g)

Exercise \(\PageIndex{b}\)

Which of the following is a Brønsted-Lowry acid?

1. (CH₃)₃NH⁺
2. CH₃COOH
3. HF
4. HNO₂
5. all of these

Answer

e. all of these

Exercise \(\PageIndex{c}\)

Which one is more acidic, HNO₂ or HNO₃?

Answer

HNO₃

Exercise \(\PageIndex{d}\)

Which one is more acidic H₃AsO₃ or H₃AsO₄?

Answer

H₃AsO₄

Exercise \(\PageIndex{e}\)

List the acids in order of increasing acid strength: HClO₂, HBrO₂, HIO₂

Answer

HIO₂< HBrO₂< HClO₂

Exercise \(\PageIndex{f}\)

List the compounds in order of increasing acid strength: AsH_3, HBr, NaH, H₂O

Answer

NaH< AsH₃< H₂O< HBr

Exercise \(\PageIndex{g}\)

List the compounds in order of increasing acid strength: H₂TeO₃, H₂TeO₄, H₂O

Answer

H₂O< H₂TeO₃< H₂TeO₄

Exercise \(\PageIndex{h}\)

What is the conjugate acid of NH₃?

Answer

NH₄⁺

Exercise \(\PageIndex{i}\)

What is the conjugate base of OH^-?

Answer

O^2-

Exercise \(\PageIndex{j}\)

The following acids are listed in order of increasing strength. List their conjugate bases in order of increasing strength. HCN, CH₃COOH, HF, HClO₄:

Answer

ClO₄^-< F^-< CH₃COO^-< CN^-

Exercise \(\PageIndex{k}\)

The hydride ion, (H^-), is a stronger base that the hydroxide ion, (OH^-). The products of the reaction H^-(aq) + H₂O(l) → products are _____.

Answer

OH^-(aq) + H₂(aq)

Exercise \(\PageIndex{l}\)

The magnitude of Kw indicated that

1. water autoionizes very slowly
2. water autoionizes very quickly
3. water autoionizes only to a very small extent
4. the autoionization of water is exothermic
5. the autoionization of water is endothermic

Answer

c. water autoionizes only to a very small extent

Exercise \(\PageIndex{m}\)

What is the concentration of water in pure water?

1. 18 M
2. 100 M
3. 55 M
4. 0.100 M
5. 83 M

Answer

c. 55 M

Exercise \(\PageIndex{n}\)

What is the pH of a solution in which the molar concentration of HCl is 1.3 * 10^-11?

Answer

\[pH = -log\left [ H^{+} \right ]=-log\left ( 1.3*10^{-11} \right )=10.89\nonumber\]

Exercise \(\PageIndex{o}\)

What is the pH of a 0.015 M solution of barium hydroxide?

Answer

\[Ba(OH)_{2}\rightarrow Ba^{2+} + 2OH^{-}\nonumber\]

\[[OH^{-}]=2*M=2*0.015M=0.03M\nonumber\]

\[pOH=-log[OH^{-}]=-log(0.03M)=1.52\nonumber\]

\[pH=14-pOH=14-1.52=12.48\nonumber\]

Exercise \(\PageIndex{p}\)

What is the pH of an aqueous solution at 25°C in which [H⁺] is 0.0025 M?

Answer

\[pH=-log\left [ H^{+} \right ]=-log\left ( 0.0025M \right )=2.60\nonumber\]

Exercise \(\PageIndex{q}\)

What is the pH of a solution that contains 3.98 * 10^-9 M hydronium ion at 25°C?

Answer

\[pH=-log\left [ H^{+} \right ]=-log\left ( 3.98*10^{-9}M \right )=8.400\nonumber\]

Exercise \(\PageIndex{r}\)

Calculate the pOH of a solution at 25°C that contains 1.94 * 10^-10M hydronium ions.

Answer

\[pH=-log\left [ H^{+} \right ]=-log\left ( 1.94*10^{-10}M \right )=9.712 \nonumber\]

\[pOH=14-pH=14-9.712=4.288\nonumber\]

Exercise \(\PageIndex{s}\)

Which solution below has the highest concentration of hydroxide ions?

1. pH = 3.21
2. pH = 12.59
3. pH = 7.93
4. pH = 9.82
5. pH = 7.00

Answer

pH = 12.59

\([OH^{-}]=10^{-12.59}=2.570*10^{-13}M \nonumber\)

Exercise \(\PageIndex{t}\)

What is the [H⁺] of an aqueous solution whose pH is 8.11?

Answer

\([H^{+}]=10^{-8.11}=7.76*10^{-9}M \nonumber\)

Exercise \(\PageIndex{u}\)

What is the concentration of H⁺ in a solution at 25°C with a pH of 7.35?

Answer

\([H^{+}]=10^{-7.35}=4.47*10^{-8}M \nonumber\)

Exercise \(\PageIndex{v}\)

What is the [OH^-] and pH of a 0.035M KOH solution at 25°C?

Answer

\[\left [KOH\right ]=\left [OH^{-}\right ]=0.035M\nonumber\]

\[pH=14-pOH =14-(-log [OH^{-}])\nonumber\]

\[14-1.46=12.54\nonumber\]

Exercise \(\PageIndex{w}\)

The pH of a 0.011 M NaOH solution at 25°C is _____.

Answer

\[pOH=-log\left (OH^{-}\right )=-log\left (0.011M\right )=1.96\nonumber\]

\[pH=14-pOH=14-1.96=12.04\nonumber\]

Exercise \(\PageIndex{x}\)

Which of the following possesses the greatest concentration of hydroxide ion?

1. a solution with a pH of 3.0
2. a 1 * 10^-4 M solution of HNO₃
3. a solution with a pOH of 12.0
4. pure water
5. a 1 * 10^-3 M solution of NH₄Cl

Answer

d. pure water

Exercise \(\PageIndex{y}\)

What is the pH of a 0.053 M solution of Ca(OH)₂?

Answer

\[\left [ OH^{-} \right ]=2*M=2*0.053M=0.11M\nonumber\]

\[pOH=-log\left [ OH^{-} \right ]=-log\left ( 0.11M \right )=0.96\nonumber\]

\[pH=14-pOH=14-0.96=13.04\nonumber\]

Exercise \(\PageIndex{z}\)

The pH of a 0.030 M HCl solution at 25°C is _____.

Answer

\[pH=-log\left [ H^{+} \right ]=-log\left ( 0.030M \right )=1.52\nonumber\]

Exercise \(\PageIndex{aa}\)

What is the [H⁺] and pH of a 0.0037 M HBr solution at 25°C?

Answer

\[\left [ HBr \right ]=\left [ H^{+} \right ]=0.0037M\nonumber\]

\[pH=-log\left [ H^{+} \right ]=-log\left ( 0.0037M \right )=2.43\nonumber\]

Exercise \(\PageIndex{ab}\)

What is the conjugate base of HSO₄^-? Conjugate acid?

Answer

The conjugate base is SO₄^2-

The conjugate acid is H₂SO₄

Exercise \(\PageIndex{ac}\)

Which of the following acids is not a strong acid?

1. H₂CO₃
2. H₂SO₄
3. HNO₃
4. HClO₄
5. HCl

Answer

a. H₂CO₃

Exercise \(\PageIndex{ad}\)

What molar concentration of aqueous barium hydroxide would have pH = 12.25?

Answer

\[pOH=14-pH=14-12.25=1.75\nonumber\]

\[\left [ OH^{-} \right ]=10^{-1.75}=0.01778M\nonumber\]

\[\left [ Ba\left ( OH \right )_{2} \right ]=\frac{\left [ OH^{-} \right ]}{2}=\frac{0.01778M}{2}=0.008891M\nonumber\]

Exercise \(\PageIndex{ae}\)

What molar concentration of aqueous hydrochloric acid would have a pH = 9.50?

Answer

It is not possible for a solution of hydrochloric acid to have a pH = 9.50

Exercise \(\PageIndex{af}\)

Which one of the following is the weakest acid

1. HF (K_a = 6.8 * 10^-4)
2. HClO (K_a = 3.0 * 10^-8)
3. HNO2 (K_a = 4.5 *10^-4)
4. HCN (K_a = 4.9 * 10^-10)
5. Acetic Acid (K_a = 1.8 *10^-5)

Answer

d. HCN (K_a = 4.9 * 10^-10)

Exercise \(\PageIndex{ag}\)

Using the table below, which is the strongest acid?

Acid	Ka
HOAc	1.8 * 10^-5
HCHO2	1.8 * 10^-4
HClO	3.0 * 10^-8
HF	6.8 * 10^-4

Answer

HF with a  K_a=6.8 * 10^-4

Exercise \(\PageIndex{ah}\)

What is the percent ionization of hypochlorous acid (HClO) in a 0.015 M aqueous solution of HClO at 25C? (K_a = 3.0 * 10^-8)

Answer

R	HClO	⇌	H^+     +	ClO^-
I	0.015M		0	0
C	-x		+x	+x
E	0.015-x		x	x

\[Ka=\frac{\left [ H^{+} \right ]\left [ ClO^{-} \right ]}{\left [ HClO \right ]}\nonumber\]

\[3.0*10^{-8}=\frac{\left [ x \right ]\left [ x \right ]}{\left [ 0.015-x \right ]}\nonumber\]

\[x=2.12*10^{-5}M \%\,ionization=\frac{Amount\,of\,acid\,ionized}{Amount\,of\,initial\,acid}*100\nonumber\]

\[\%\,ionization=\frac{2.12*10^{-5}}{0.015}*100=0.14\%\ \nonumber\]

Exercise \(\PageIndex{ai}\)

A 0.15M aqueous solution of the weak acid HA at 25°C has a pH of 5.35. What is the value of K_a for HA?

Answer

R	HA	⇌	H^+     +	A-
I	0.15M		0	0
C	-x		+x	+x
E	0.15-x		x	x

\[Ka=\frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{(x)(x)}{(0.15-x)} \nonumber \]

\[\left [ H^{+} \right ]=x=10^{-pH}=10^{-5.35}=4.47*10^{-6}M \nonumber \]

\[Ka=\frac{(x)(x)}{(0.15-x)}=\frac{(4.47*10^{-6})(4.47*10^{-6})}{(0.15-(4.47*10^{-6}))}=\frac{2.00*10^{-11}}{0.150}=1.33*10^{-10} \nonumber \]

Exercise \(\PageIndex{aj}\)

The K_a of HClO is 3.0*10^-8. What is the pH at 25°C of an aqueous solution that is 0.020M in HClO?

Answer

R	HClO	⇌	H^+     +	ClO^-
I	0.020M		0	0
C	-x		+x	+x
E	0.020-x		x	x

\[Ka=\frac{\left [ H^{+} \right ]\left [ ClO^{-} \right ]}{\left [ HClO \right ]}=\frac{(x)(x)}{(0.020-x)} \nonumber \]

\[3.0*10^{-8}=\frac{(x)(x)}{(0.020-x)} \nonumber \]

\[\left [ H^{+} \right ]=x=2.5*10^{-5}M \nonumber \]

\[pH=-log\left [ H^{+} \right ]=-log\left ( 2.5*10^{-5} \right )=4.61 \nonumber \]

Exercise \(\PageIndex{ak}\)

The K_a of HF is 6.8*10^-4. What is the pH of a 0.35M solution of HF? 1.81

Answer

R	HF	⇌	H^+     +	F^-
I	0.35M		0	0
C	-x		+x	+x
E	0.35-x		x	x

\[Ka=\frac{\left [ H^{+} \right ]\left [ F^{-} \right ]}{\left [ H \right ]}=\frac{(x)(x)}{(0.35-x)} \nonumber \]

\[6.8*10^{-4}=\frac{(x)(x)}{(0.35-x)} \nonumber \]

\[\left [ H^{+} \right ]=x=1.5*10^{-2}M \nonumber \]

\[pH=-log\left [ H^{+} \right ]=-log\left ( 1.5*10^{-2} \right )=1.81 \nonumber \]

Exercise \(\PageIndex{al}\)

A 0.25M solution of the weak acid HX has a pH of 4.15. What is the value of K_a for HX? 

Answer

\[HX\rightleftharpoons H^{+}+X^{-} \nonumber \]

\[Ka=\frac{\left [ H^{+} \right ]\left [ X^{-} \right ]}{\left [ HX \right ]} \nonumber \]

\[\left [ H^{+} \right ]=\left [ A^{-} \right ]=10^{-pH}=10^{-4.15}=7.08*10^{-5}M \nonumber \]

\[Ka=\frac{\left [ H^{+} \right ]\left [ X^{-} \right ]}{\left [ HX \right ]}=\frac{\left ( 7.08*10^{-5}M \right )\left ( 7.08*10^{-5}M \right )}{\left ( 0.25M \right )}=2.0*10^{-8} \nonumber \]

Exercise \(\PageIndex{am}\)

A 0.0035M aqueous solution of a compound has a pH=2.46. The compound is

1. a weak base
2. a weak acid
3. a strong acid
4. a strong base
5. a salt

Answer

c. a strong acid

Exercise \(\PageIndex{an}\)

In which of the following aqueous solution does the weak acid exhibit the highest percentage ionization?

1. 0.01M HC₂H₂C₂ (K_a =3.0*10-⁸)
2. 0.01M HNO₂ (K_a = 4.5*10^-4)
3. 0.01M HF (K_a = 6.8*10^-4)
4. 0.01M HClO (K_a = 3.0*10^-8)
5. These will all exhibit the same percentage ionization

Answer

c. 0.01M HF (K_a = 6.8*10^-4)

Exercise \(\PageIndex{ao}\)

An aqueous solution of phosphoric acid has a concentration of 2.5M. (Ka₁ = 7.5*10^-3, Ka₂ = 6.2*10^-8, Ka₃ = 4.2*10^-13)

1. What is the pH?
2. What is the molar concentration of phosphate ion?

Answer a.

First dissociation of phosphoric acid:

\[H_{3}PO_{4}(aq)\rightleftharpoons H^{+}(aq)+H_{2}PO_{4}^{-}(aq)\,\,\,K_{a1}=7.5*10^{-3} \nonumber \]

\[H_{2}PO_{4}^{-}\rightleftharpoons H^{+}(aq)+HPO_{4}^{2-}(aq)\,\,\,K_{a2}=6.2*10^{-8} \nonumber \]

\[HPO_{4}^{2-}(aq)\rightleftharpoons H^{+}(aq)+PO_{4}^{3-}(aq)\,\,\,K_{a3}=4.2*10^{-13} \nonumber \]

R	H3PO4(aq)	⇌	H^+(aq)     +	H2PO4^-(aq)
I	2.5M		0	0
C	-x		+x	+x
E	2.5+x		x	x

\[K_{a1}=\frac{\left [H^{+}\right ]\left [H_{2}PO_{4}^{-}\right ]}{\left [H_{3}PO_{4}\right ]}=\frac{(x)(x)}{(2.5-x)}=7.5*10^{-3} \nonumber \]

\[x=\left [H^{+}\right ]=0.13323M \nonumber \]

 

Second dissociation of phosphoric acid:

R	H2PO4^-(aq)	⇌	H^+(aq)     +	HPO4^2-(aq)
I	0.13323M		0.13323M	0
C	-y		+y	+y
E	0.13323-y		0.13323+y	y

\[K_{a2}=\frac{\left [ H^{+} \right ]\left [ HPO_{4}^{2-} \right ]}{\left [ H_{2}PO_{4}^{-} \right ]}=\frac{(y)(0.13323+y)}{0.13323-y}=6.2*10^{-8} \nonumber \]

(Assume <<0.13323 M)

\[\left [ H^{+} \right ]=K_{a2}=6.2*10^{-8} \nonumber \]

The [H+] from the second step is negligible.

 

Third dissociation of phosphoric acid:

R	HPO4^2-(aq)	⇌	H^+(aq)     +	PO4^3-(aq)
I	6.2*10-8M		0.13323M	0
C	-z		+z	+z
E	6.2*10^-8-z		0.13323+z	y

\[K_{a3}=\frac{\left [ H^{+} \right ]\left [ PO_{4}^{3-} \right ]}{\left [ HPO_{4}^{2-} \right ]}=\frac{\left ( z \right )\left ( 0.13323+z \right )}{\left ( 6.2*10^{-8} \right )-z}=4.2*10^{-13} \nonumber \]

\[z=1.4*10^{-19}M \nonumber \]

The [H+] from the third step is negligible.

\[pH=-log\left [ H^{+} \right ]=-log\left [ 0.13323 \right ]=0.88 \nonumber \]

Answer b.

First dissociation of phosphoric acid:

R	H3PO4(aq)	⇌	H^+(aq)     +	H2PO4^-(aq)
I	2.5M		0	0
C	-x		+x	+x
E	2.5+x		x	x

\[K_{a1}=\frac{\left [H^{+}\right ]\left [H_{2}PO_{4}^{-}\right ]}{\left [H_{3}PO_{4}\right ]}=\frac{(x)(x)}{(2.5-x)}=7.5*10^{-3} \nonumber \]

\[x=\left [H_{2}PO_{4}^{-}\right ]=0.13323M \nonumber \]

 

For the second dissociation:

\[K_{a2}=\frac{\left [ H^{+} \right ]\left [ HPO_{4}^{2-} \right ]}{\left [ H_{2}PO_{4}^{-} \right ]}=\frac{(x)(x)}{(0.13323-x)}=6.2*10^{-8} \nonumber \]

\[x=\left [ HPO_{4}^{2-} \right ]=1.92*10^{-4}M \nonumber \]

 

For the third dissociation:

\[K_{a3}=\frac{\left [ H^{+} \right ]\left [ PO_{4}^{3-} \right ]}{\left [ HPO_{4}^{2-} \right ]}=\frac{(x)(x)}{(1.92*10^{-4})-x}=4.2*10^{-13} \nonumber \]

\[x=\left [ PO_{4}^{3-} \right ]=1.92*10^{-19}M \nonumber \]

Exercise \(\PageIndex{ap}\)

Which species form the following list would be the strongest Bronsted-Lowry base?

1. Cl^-
2. Br^-
3. NO₃^-
4. F^-
5. ClO^-

Answer

d. F^-

Exercise \(\PageIndex{aq}\)

Which of the following ions will act as a weak base in water?

1. OH^-
2. Cl^-
3. NO₃^-
4. ClO^-
5. None of these will act as a weak base in water

Answer

d. ClO^-

Exercise \(\PageIndex{ar}\)

The pH of a 0.10M solution of a weak base is 9.82. What is the Kb for this base?

Answer

\[pH+pOH=14 \nonumber \]

\[pOH=14-9.82=4.18 \nonumber \]

\[ [OH^{-}] = 10^{-pOH} = 10^{-4.18}= 6.61*10^{5} M \nonumber \]

\[K_{b}=\frac{\left [ B^{+} \right ]\left [ OH^{-} \right ]}{\left [ BOH \right ]}=\frac{\left ( 6.61*10^{5} \right )^{2}}{\left ( 0.1-\left ( 6.61*10^{5} \right ) \right )}=4.37*10^{-8} \nonumber \]

Exercise \(\PageIndex{as}\)

Given that the Ka for gallic acid, (HC₇H₅O₅) is 4.57*10^-3, what is the Kb for the gallate ion (NaC₇H₅O₅)? T = 25°C

Answer

\[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{4.57*10^{-3}}=2.19*10^{-12} \nonumber \]

Exercise \(\PageIndex{at}\)

Kb for C₅H₅N is 1.4*10^-9. Ka for C₅H₅NH⁺ is _____. T = 25°C

Answer

\[K_{a}=\frac{K_{w}}{K_{b}}=\frac{10^{-14}}{1.4*10^{-9}}=7.1*10^{-6} \nonumber \]

Exercise \(\PageIndex{au}\)

Ka for HF is 7.0*10^-4. Kb for the fluoride ion is _____.

Answer

\[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{7.0*10^{-4}}=1.4*10^{-11} \nonumber \]

Exercise \(\PageIndex{av}\)

Calculate the pOH of a 0.0827M aqueous sodium cyanide solution at 25°C (for CN^-, Kb = 4.9*10^-10).

Answer

R	CN^-(aq)     +	H2O(aq)	⇌	HCN(aq)     +	OH^-(aq)
I	0.0827M			0	0
C	-x			+x	+x
E	0.0827-x			x	x

\[K_{b}=\frac{\left [ HCN \right ]\left [ OH^{-} \right ]}{\left [ CN^{-} \right ]}=\frac{x^{2}}{0.0827-x}=4.9*10^{-10} \nonumber \]

\[x=\left [ OH^{-} \right ]=6.4*10^{-6}M \nonumber \]

\[pOH=-log\left [ OH^{-} \right ]=-log\left [ 6.4*10^{-6}M \right ]=5.20 \nonumber \]

Exercise \(\PageIndex{aw}\)

Determine the pH of a 0.15M solution of KF. For hydrofluoric acid, Ka = 7.0*10^-4.

Answer

\[K_{b}=\frac{K_{w}}{K_{a}}\frac{10^{-14}}{7.0*10^{-4}}=1.4*10^{-11} \nonumber \]

\[K_{b}=\frac{\left [ HF \right ]\left [ OH^{-} \right ]}{\left [ F^{-} \right ]}=\frac{(x)(x)}{0.15-x}=1.4*10^{-11} \nonumber \]

\[x=\left [ OH^{-} \right ]=1.45*10^{-6}M \nonumber \]

\[pOH=-log\left [ OH^{-} \right ]=-log\left [ 1.45*10^{-6}M \right ]=5.84 \nonumber \]

\[pH=14-pOH=14-5.84=8.16 \nonumber \]

Exercise \(\PageIndex{ax}\)

Calculate the pH of 0.726M anilinium hydrochloride, (C₆H₅NH₃Cl) solution in water given that Kb for aniline is 3.83*10^-4.

Answer

\[K_{a}=\frac{K_{w}}{K_{b}}\frac{10^{-14}}{3.83*10^{-4}}=2.61*10^{-11} \nonumber \]

\[K_{a}=\frac{\left [ C_{6}H_{5}NH_{2} \right ]\left [ H^{+} \right ]}{\left [ C_{6}H_{5}NH_{3}^{+} \right ]}=\frac{(x)(x)}{0.726-x}=2.61*10^{-11} \nonumber \]

\[x=\left [ H^{+} \right ]=4.35*10^{-6}M pH=-log\left [ H^{+} \right ]=-log\left [ 4.35*10^{-6}M \right ]=5.36 \nonumber \]

Exercise \(\PageIndex{ay}\)

The Ka for formic acid (HCHO₂) is 1.8*10^-4. What is the pH of a 0.35M solution of sodium formate (NaCHO₂)?

Answer

R	HCOO^-(aq)     +	H2O(l)		HCOOH(aq)     +	OH^-(aq)
I	0.35M			0	0
C	-x			+x	+x
E	0.35-x			x	x

\[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{1.8*10^{-4}}=5.56*10^{-11} \nonumber \]

\[K_{b}=\frac{\left [ HCOOH \right ]\left [ OH^{-} \right ]}{\left [ HCOO^{-} \right ]}=\frac{(x)(x)}{0.35-x}=5.56*10^{-11} \nonumber \]

\[x=\left [ OH^{-} \right ]=4.41*10^{-6}M \nonumber \]

\[pOH=-log\left [ OH^{-} \right ]=-log\left [ 4.41*10^{-6}M \right ]=5.36 \nonumber \]

\[pH=14-pOH=14-5.36=8.64 \nonumber \]

Exercise \(\PageIndex{az}\)

A 0.1M aqueous solution of _____ will have the highest pH.

1. KCN, Ka of HCN = 4.0*10^-10
2. NH₂NO₃, Kb of NH₃ = 1.8*10^-5
3. NaOAc, Ka of HOAc = 1.8*10^-5
4. NaClO, Ka of HClO = 3.2*10^-8
5. NaHS, Kb of HS^- = 1.8*10^-7

Answer

a. KCN, Ka of HCN = 4.0*10^-10

Exercise \(\PageIndex{ba}\)

A 0.1M solution of _____ has a pH of 7.0.

1. Na₂S
2. KF
3. NaNO₃
4. NH₃Cl
5. NaF

Answer

c. NaNO₃

Exercise \(\PageIndex{bb}\)

Ka of HX is 7.5*10^-12. What is the pH of a 0.15M solution of NaX?

Answer

\[ X^{-} +H_{2}O \leftrightarrow HX + OH^{-} \nonumber \]

\[Kb = \frac {[HX][OH^{-}]}{[X^{-}]}\nonumber\]

\[Kb= \frac {Kw}{Ka}\nonumber\]

\[Kb=\frac {1*10^{-14}}{7.5*10^{-12}} = 1.33*10^{-3} \nonumber\]

\[1.33*10^{-3} =\frac {x^{2}}{0.15-x} \nonumber\]

Assume that \(x\) is very small in denominator

\[1.33*10^{-3} =\frac {x^{2}}{0.15} \nonumber\]

\[x=0.01412 = [OH^{-}]\nonumber\]

\[pOH=-log [OH^{-}]\nonumber\]

\[pOH=1.85\nonumber\]

\[pH = 14-pOH = 12.15\nonumber\]

Exercise \(\PageIndex{bc}\)

Of the following, which is the strongest acid?

1. HIO
2. HIO₄
3. HIO₂
4. HIO₃
5. The acid strength of all these is nearly the same.

Answer

b. HIO₄

Exercise \(\PageIndex{bd}\)

Of the following, the acid strength of _____ is the greatest.

1. CH₃COOH
2. ClCH₂COOH
3. Cl₂CHCOOH
4. Cl₃CCOOH
5. BrCH₂COOH

Answer

d. Cl₃CCOOH

Exercise \(\PageIndex{be}\)

Of the following, _____ is the strongest acid.

1. Cl₃C-COOH
2. H₃C-COOH
3. Br₂C-COOH
4. F₃C-COOH
5. Br₂ClC-COOH

Answer

d.F₃C-COOH

Exercise \(\PageIndex{bf}\)

Which of the following acids will be the strongest?

1. H₂SO₄
2. HSO₄^-
3. H₂SO₃
4. H₂SeO₄
5. HSO₃^-

Answer

a.  H₂SO₄

Exercise \(\PageIndex{bg}\)

The more electronegative X is, the _____ polar will be the H-X bond and the _____ easily the H-X bond is broken, making HX more _____ acidic.

1. more, less, weakly
2. more, more, weakly
3. more, more, strongly
4. more, less, strongly
5. less, less, strongly

Answer

c. more, more, strongly

Exercise \(\PageIndex{bh}\)

Which one of the following cannot act as a Lewis base?

1. Cl^-
2. NH₃
3. BF₃
4. CN^-
5. H₂O

Answer

C. BF₃

Textbook: 16: Acids and Bases

Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, [e-mail]. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

• Liliane Poirot