16: Acids and Bases

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16.1: Brønsted-Lowry Concept of Acids and Bases

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

H₂O, NH₃

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

What is the conjugate acid of NH₃?

Answer

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

base            acid            C.A.            C.B.

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

What is the conjugate base of OH^-?

Answer

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

base            acid            C.A.            C.B.

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

What is the conjugate acid of NaHSO₃?

Answer

Amphoteric Substance

\[HSO_{3}^{-}+H^{+}\rightarrow H_{2}CO_{3} \]

Acid salts are amphoteric.

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

What is the conjugate base of NaHSO₃?

Answer

Amphoteric Substance

\[HSO_{3}^{-}+OH^{-}\rightarrow SO_{3}^{-2}+H_{2}O\]

Acid salts are amphoteric.

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

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

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

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

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

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

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

Answer

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

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

What is [OH^-] of the detergent in Q 16.2.2?

Answer

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

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

What is the pH of 0.055M of HCl?

Answer

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

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

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

What is the pH of 0.001M of Ca(OH)₂?

Answer

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

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

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

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

What is the pOH of the solution in Q 16.2.5?

Answer

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

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

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

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

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

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

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

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

 

16.3: Equilibrium Constants for Acids and Bases

Weak Acids

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

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

Answer

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

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

Therefore,

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

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

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.

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

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

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

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

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

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

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

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

Answer

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

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

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

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

Answer

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

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

Therefore,

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

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

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

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

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

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

 

Weak Base

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

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

Answer

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[pOH=2.90 \]

\[pH=14.00-2.90=11.10\]

 

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

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

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

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

\[pH=4.53\]

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

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

Answer

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

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

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

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

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

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

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

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

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

\[pH=1.66\]

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

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

Answer

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

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

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

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

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

Answer

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

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

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

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

 

16.4: Acid-Base Properties of Salts

Percent Ionization

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

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

Answer

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

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

Therefore,

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

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

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

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

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

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

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

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

Answer

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

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

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

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

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

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

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

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

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

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

Answer

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

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

Therefore,

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

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

 

pH of Various Salts

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

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

Answer

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

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

Therefore,

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

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

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

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

Answer

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

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

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

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

\[pH=14.00-3.89=10.11\]

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

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

Answer

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

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

Therefore,

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

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

\[pH=14.00-4.47=9.53\]

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

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

Answer

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

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

Therefore,

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

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

\[pH=14.00-6.06=7.94\]

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

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

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

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

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

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

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

 

16.5: Acid-Base Equilibrium Calculations

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

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

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

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

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

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

Answer

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

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

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

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

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

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

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

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

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

\[\left [H^{+}\right ]=\sqrt{4.3*10^{-7}*0.002}=2.93*10^{-5}M \[pH=-log\left (2.93*10^{-5}M\right )=4.53\]

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

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

Answer

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

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

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

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

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

\[pH=4.50\]

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

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

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

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

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

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

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

 

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

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

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

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

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

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

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

Answer

\[pH=13.0\]

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

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

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

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

\[x=0.05mol\]

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

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

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

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

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

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

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

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

Answer

\[pH=13.0\]

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

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

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

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

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

\[x=0.05mol\]

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

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

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

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

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

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

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

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

 

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

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

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

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

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

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

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

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

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

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

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

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

 

16.7: Lewis Concept of Acids and Bases

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

Answer

Lewis base

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

In Q 16.7.1, Is the metal, such as Fe(II), a lewis acid or a lewis base?

Answer

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

\(Zn\left ( OH \right )_{2}\)

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

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

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

Exercise \(\PageIndex{o}\)

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

Answer

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

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

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

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

Exercise \(\PageIndex{p}\)

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

Answer

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

Exercise \(\PageIndex{q}\)

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

Answer

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

Exercise \(\PageIndex{r}\)

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

Answer

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

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

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

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

Exercise \(\PageIndex{t}\)

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

Answer

\[H⁺]=10^{-8.11}=7.76*10^{-9}M\]

Exercise \(\PageIndex{u}\)

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

Answer

\[H⁺]=10^{-7.35}=4.47*10^{-8}M\]

Exercise \(\PageIndex{v}\)

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

Answer

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

\[pH=14-pOH=14-\left ( -log\left [OH^{-}\right ] \right 14-1.46=12.54\]

Exercise \(\PageIndex{w}\)

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

Answer

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

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

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

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

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

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

Exercise \(\PageIndex{z}\)

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

Answer

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

Exercise \(\PageIndex{aa}\)

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

Answer

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

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

Exercise \(\PageIndex{ab}\)

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

Answer

SO₄^2-, 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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:

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

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

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

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

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

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

 

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

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

(Assume <<0.13323 M)

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

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

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

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

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

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

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

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

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

 

For the second dissociation:

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

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

 

For the third dissociation:

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

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

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

\[pH+pOH=14\]

\[pOH=14-9.82=4.18\]

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

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

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

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

Exercise \(\PageIndex{at}\)

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

Answer

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

Exercise \(\PageIndex{au}\)

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

Answer

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

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

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

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

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

Exercise \(\PageIndex{aw}\)

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

Answer

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Add texts here. Do not delete this text first.

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