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17: Aqueous Equilibria

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    17.2: Controlling pH: Buffer Solutions

    Textbook: Section 17.2

    Buffer Solutions

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

    The Henderson-Hasselbalch equation is

    1. \(\left [ H^{+} \right ]=K_{a}+\frac{\left [ base \right ]}{\left [ acid \right ]}\)
    2. \(pH=pK_{a}-log\frac{\left [ base \right ]}{\left [ acid \right ]}\)
    3. \(pH=pK_{a}+log\frac{\left [ base \right ]}{\left [ acid \right ]}\)
    4. \(pH=pK_{a}+log\frac{\left [ acid \right ]}{\left [ base \right ]}\)
    5. \(pH=log\frac{\left [ acid \right ]}{\left [ base \right ]}\)
    Answer

    c. \(pH=pK_{a}+log\frac{\left [ base \right ]}{\left [ acid \right ]}\)

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

    What is the pH of a solution formed by adding 0.20mol HF and 0.20mol of NaF to enough water to make 1.0L of the solution? Ka=6.8x10-4

    Answer
    R HF(aq) H+(aq)     + F-(aq)
    I 0.20 M   0 0.20 M
    C -x   +x +x
    E 0.20-x   x 0.20+x

    \[K_{a}=\frac{x\left ( 0.20+x \right )}{0.20-x}=6.8*10^{-4}\nonumber\]

    \[x=\left [ H^{+} \right ]=6.8*10^{-4}\nonumber\]

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 6.8*10^{-4} \right )=3.17\nonumber\]

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

    What is the pH of a solution formed by 0.3M HF and 0.1M HCl?

    Answer
    R HF(aq) H+(aq)     + F-(aq)
    I 0.30 M   0.10 M 0
    C -x   +x +x
    E 0.30-x   0.10+x x

    \[K_{a}=\frac{x\left ( 0.10+x \right )}{0.30-x}=6.8*10^{-4}\nonumber\]

    \[x=0.002 M\nonumber\]

    \[\left [ H^{+} \right ]=0.10+0.002=0.102M\nonumber\]

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 0.102 M \right )=1.0\nonumber\]

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

    What is the pH of the solution if 0.020mol of NaOH is added to the 1.0L solution formed by adding 0.200mol of HF and 0.200mol of NaF?

    Answer

    After the reaction,

    HF(aq)     + OH-(aq) H2O(l)     + F-(aq)
    0.180 M 0     0.220

    pH before the reaction, from Q 17.2.b, is 3.17.

    pH after the reaction

    \[pH=3.17+log\frac{0.220}{0.180}=3.26\nonumber\]

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

    What is the ratio of HCO3- to H2CO3 in our blood that has a pH=7.4? Ka1=4.3x10-7

    Answer

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

    \[K_{a1}=\frac{\left [ H^{+} \right ]\left [ HCO_{3}^{-} \right ]}{\left [ H_{2}CO_{3} \right ]}\nonumber\]

    \[\frac{\left [ HCO_{3}^{-} \right ]}{\left [ H_{2}CO_{3} \right ]}=\frac{K_{a1}}{\left [ H^{+} \right ]}=\frac{4.3*10^{-7}}{10^{-7.4}}=10.8\nonumber\]

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

    How many grams of NH4Cl must be added to 1.00L of 0.10M NH3 to form a buffer that has a pH=9.0?  Kb=1.8x10-5 (Regardless the volume change.)

    Answer

    \[\left [ OH^{-} \right ]=10^{-\left ( 14-9 \right )}=10^{-5}M \nonumber \]

    R NH3(l)     + H2O(l) NH4+(aq)     + OH-(aq)
    I 0.10 M ---   A 0
    C -x ---   +x +x
    E 0.10-x ---   A+x x

    \[K_{b}=\frac{\left [ OH^{-} \right ]\left [ NH_{4}^{+} \right ]}{\left [ NH_{3} \right ]}=\frac{10^{-5}\left ( A+10^{5} \right )}{0.10-10^{-5}}=1.8*10^{-5} \nonumber \]

    \[A=0.18M \nonumber \]

    \[0.18mol*53.45g/mol=9.62g \nonumber \]

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

    Which one of the following pairs cannot be mixed to form a buffer solution?

    1. H3PO4, KH2PO4
    2. KOH, HF
    3. NaC2H3O2, HCl
    4. NH3, NH4Cl
    5. RbOH, HBr
    Answer

    e. RbOH, HBr

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

    Consider a solution containing 0.100M fluoride ion and 0.126M hydrogen fluoride. The concentration of hydrogen fluoride after addition of 5.00mL if 0.0100M HCl to 25.0mL of this solution is ______.

    Answer

    \[mol_{F^{-}}= 0.100M*0.0250L=2.50*10^{-3}mol \nonumber \]

    \[mol_{HF}=0.126M*0.0250L=3.15*10^{-3}mol \nonumber \]

    \[mol_{HCl}=mol_{H^{+}}=0.0100M*0.00500L=5.00*10^{-5}mol \nonumber \]

    \[x=mol_{H^{+}}=5.00*10^{-5}mol \nonumber \]

    R F-(aq)     + H+(aq) HF(aq)
    I 2.50*10-3 mol 5.00*10-5 mol   3.15*10-3 mol
    C -x -x   +x
    E 2.50*10-3 - x 0   3.15*10-3 + x

    \[F^{-}=\left ( 2.50*10^{-3}mol \right )-\left ( 5.00*10^{-5}mol \right )=2.45*10^{-3}mol \nonumber \]

    \[HF=\left ( 3.15*10^{-3}mol \right )+\left ( 5.00*10^{-5}mol \right )=3.20*10^{-3}mol \nonumber \]

    \[\left [ HF \right ]=\frac{3.20*10^{-3}mol}{0.0250L+0.00500L}=0.107M \nonumber \]

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

    Of the following, which solution has the greatest buffering capacity?

    1. 0.821 M HF and 0.217 M NaF
    2. 0.821 M HF and 0.909 M NaF
    3. 0.100 M HF and 0.217 M NaF
    4. 0.121 M HF and 0.667 M NaF
    5. They are all buffer solutions and would all have the same capacity.
    Answer

    b. 0.821 M HF and 0.909 M NaF

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

    The Ka of acetic acid is 1.7*10-5. What is the pH of a buffer prepared by combining 50.0mL of 1.00M potassium acetate and 50.0mL of 1.00M acetic acid?

    Answer

    \[\left [ CH_{3}COOK \right ]=1.00M*0.0500L=0.0500mol \nonumber \]

    \[\left [ CH_{3}COOH \right ]=1.00M*0.0500L=0.0500mol \nonumber \]

    \[pOH=pK_{a}+log\frac{\left [ CH_{3}COOK \right ]}{\left [ CH_{3}COOH \right ]}=-log\left ( 1.7*10^{-5} \right )+log\frac{0.0500mol}{0.0500mol}=4.77 \nonumber \]

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

    The Kb of ammonia is 1.8*10-5. What is the pH of a buffer prepared by combining 50.0mL of 1.00M ammonia and 50.0mL of 1.00M ammonium nitrate?

    Answer

    \[\left [ NH_{4}NO_{3} \right ]=1.00M*0.0500L=0.0500mol \nonumber \]

    \[\left [ NH_{3} \right ]=1.00M*0.0500L=0.0500mol \nonumber \]

    \[pOH=pK_{b}+log\frac{\left [ NH_{4}NO_{3} \right ]}{\left [ NH_{3} \right ]}=-log\left ( 1.8*10^{-5} \right )+log\frac{0.0500mol}{0.0500mol}=4.77 \nonumber \]

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

    Calculate the pH of a solution prepared by dissolving 0.25 mol of benzoic acid (C7H5O2H) and 0.15 mol of sodium benzoate (NaC7H5O2) in 1.00L of solution. Ka = 6.5*10-5 for benzoic acid.

    Answer

    \[pH=pK_{a}+log\frac{salt}{acid}=-log(6.5*10^{-5})+log\frac{0.150}{0.250}=4.19+(-0.22)=3.97 \nonumber \]

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

    Determine the pH of a solution prepared by adding 0.45 mol of solid KOAc to 1.00L of 2.00M HOAc. Ka = 1.8*10-5 of HOAc.

    Answer

    \[pH=pK_{a}+log\frac{\left [ CH_{3}COO^{-} \right ]}{\left [ CH_{3}COOH \right ]}=-log\left ( 1.8*10^{-5} \right )+log\frac{0.45mol}{2.00mol}=4.10 \nonumber \]

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

    The pH of a solution is prepared by dissolving 0.35 mol of solid CH3NH3Cl (methylamine hydrochloride) in 1.00L of 1.10 M CH3NH2 (methylamine) is _____. The Kb for methylamine is 4.4*10-4.

    Answer

    \[pOH=pK_{b}+log\frac{\left [ CH_{3}NH_{3}^{+} \right ]}{\left [ CH_{3}NH_{2} \right ]}=-log\left ( 4.4*10^{-4} \right )+log\frac{0.35mol}{1.10mol}=2.86 \nonumber \]

    \[pH=14-pOH=14-2.86=11.14 \nonumber \]

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

    Which of the following substances, when added to a solution of hydrofluoric acid, could be used to prepare a buffer solution?

    1. HCl
    2. NaNO3
    3. NaF
    4. NaCl
    5. NaBr
     
    Answer

    c. NaF

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

    Which of the following could not be added to a solution of sodium acetate to prepare a buffer?

    1. ammonium acetate
    2. acetic acid
    3. hydrochloric acid
    4. nitric acid
    5. more than one of these answers is correct
    Answer

    a. ammonium acetate

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

    The primary buffer system that controls the pH of the blood is the _____ buffer system.

    1. carbon dioxide, carbonate
    2. carbonate, bicarbonate
    3. carbonic acid, carbon dioxide
    4. carbonate, carbonic acid
    5. carbonic acid, bicarbonate
    Answer

    e. carbonic acid, bicarbonate

     

    17.3: Acid-Base Titrations

    Textbook: Section 17.3

    Acid-Base Titrations

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

    How many milliliters of 0.0750M NaOH are required to titrate 50.0ml of 0.025M HCl?

    Answer

    \[OH^{-}(aq)+H^{+}\rightleftharpoons H_{2}O(l) \nonumber \]

    \[\frac{50.0mL*0.025M}{0.075M}=16.7mL \nonumber \]

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

    How many milliliters of the same NaOH are needed to titrate 25.0ml of HCl solution that has 1.85g of HCl per liter?

    Answer

    \[OH^{-}(aq)+H^{+}\rightleftharpoons H_{2}O(l) \nonumber \]

    \[\frac{\frac{1.85g/L}{36.45g/mol}*25.0mL}{0.075M}=16.9mL \nonumber \]

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

    How many milliliters of the same NaOH are needed to titrate 20.0ml of 0.050M HBr?

    Answer

    \[OH^{-}(aq)+HBr\rightleftharpoons H_{2}O(l)+Br^{-}(aq) \nonumber \]

    \[\frac{0.050M*20.0mL}{0.075M}=13.3mL \nonumber \]

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

    Calculate the pH of the solution formed when 20.0ml of 0.100M NaOH is added to 40.0ml of 0.100M of HC2H3O2. Ka=1.8x10-5

    Answer

    \[\frac{20.0mL}{1000ml/L}*0.100M=0.002mol\,NaOH \nonumber \]

    \[\frac{40.0mL}{100mL/L}*0.100M=0.004mol\,HC_{2}H_{3}O_{2} \nonumber \]

    HC2H3O2(aq)     + OH-(aq) C2H3O2-(aq)     + H2O(l)
    0.002mol left ---   0.002mol produced ---

    \[Total\,volume=20.0+40.0=60.0mL \nonumber \]

    \[\left [ HC_{2}H_{3}O_{2} \right ]=\frac{0.002mol}{0.060L}=0.0333M \nonumber \]

    \[\left [ C_{2}H_{3}O_{2}^{-} \right ]=\frac{0.002mol}{0.060L}=0.0333M \nonumber \]

    \[K_{a}=\frac{\left [ C_{2}H_{3}O_{2}^{-} \right ]\left [ H^{+} \right ]}{\left [ HC_{2}H_{3}O_{2} \right ]}=1.8*10^{-5} \nonumber \]

    \[\left [ H^{+} \right ]=1.8*10^{-5}M \nonumber \]

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 1.8*10^{-5}M \right )=4.74 \nonumber \]

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

    Calculate the pH at the equivalence point in the titration of 50.0ml of 0.20M HC2H3O2 with 0.05M NaOH. Ka=1.8x10-5

    Answer

    \[\frac{50.0mL*0.20M}{0.05}M=200.0mL \nonumber \]

    \[\left [ C_{2}H_{3}O_{2}^{-} \right ]=\frac{0.20M*0.050L}{\left ( 0.050+0.200 \right )L}=0.040M \nonumber \]

    \[C_{2}H_{3}O_{2}^{-}(aq)+H_{2}O(l)\rightleftharpoons HC_{2}H_{3}O_{2}(aq)+OH^{-}(aq) \nonumber \]

    \[K_{b}=\frac{K_{w}}{K_{a}}=5.6*10^{-5}=\frac{x^{2}}{0.040-x}=\frac{x^{2}}{0.040} \nonumber \]

    \[x=\left [ OH^{-} \right ]=4.7*10^{-6} \nonumber \]

    \[pH=-log\left [ H^{+} \right ]=-log\left ( \frac{10^{-14}}{\left [ OH^{-} \right ]} \right )=-log\left ( \frac{10^{-14}}{4.7*10^{-6}} \right )=8.68 \nonumber \]

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

    Determine the pH when 50.0ml of 0.010M NaOH is added to 40.0ml of 0.010M HC2H3O2.

    Answer

    \[\left [ OH^{-} \right ]=\frac{\left ( 0.050*0.010 \right )-\left ( 0.040*0.010 \right )mol}{0.090L}=0.0011M \nonumber \]

    \[pH=-log\left [ H^{+} \right ]=-log\left ( \frac{10^{-14}}{\left [ OH^{-} \right ]} \right )=-log\left ( \frac{10^{-14}}{0.0011M} \right )=11.05 \nonumber \]

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

    Consider the titration of 25.0mL of 0.723M HClO4 with 0.273M KOH.

    1. What is the H3O+ concentration before any KOH is added?
    2. What is the H3O+ concentration after addition of 10.0mL of KOH?
    3. What is the H3O+ concentration after addition of 66.2mL of KOH?
    4. What is the H3O+ concentration after addition of 80.0mL of KOH?
    Answer a.

    \[\left [ H_{3}O^{+} \right ]=\left [ HClO_{4} \right ]=\frac{0.723mol}{0.0250L}=0.0181M \nonumber \]

    Answer b.

    \[moles\,HClO_{4}=0.723M*0.0250L=0.018075mol \nonumber \]

    \[moles\,KOH=0.273M*0.0100L=0.00273mol \nonumber \]

    \[moles\,H_{3}O^{+}=moles\,HClO_{4}-moles\,KOH=0.018075mol-0.00273mol=0.015345mol \nonumber \]

    \[\left [ H_{3}O^{+} \right ]=\frac{0.015345mol}{0.0250L+0.0100L}=0.438M \nonumber \]

    Answer c.

    \[moles\,HClO_{4}=0.723M*0.0250L=0.018075mol \nonumber \]

    \[moles\,KOH=0.273M*0.0662L=0.0180725mol \nonumber \]

    The solution is neutral because the acid and base have the same amount of moles. So the concentration of H3O+ is 1.00*10-7.

    Answer d.

    \[moles\,HClO_{4}=0.723M*0.0250L=0.018075mol \nonumber \]

    \[moles\,KOH=0.273M*0.0800L=0.02184mol \nonumber \]

    \[excess\,moles=moles\,KOH-moles\,HClO_{4}=0.02184mol-0.018075mol=0.003765mol \nonumber \]

    \[\left [ OH^{-} \right ]=\frac{0.003765mol}{0.0800L+0.0250L}=0.0359M \nonumber \]

    \[\left [ H_{3}O^{+} \right ]=\frac{10^{-14}}{\left [ OH^{-} \right ]}=\frac{10^{-14}}{0.0359M}=2.79*10^{-13}M \nonumber \]

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

    A _____ yields a titration curve with an initial pH of 1.00, an equivalence point at pH 7.00, and a relatively long, nearly vertical middle section.

    1. strong acid titrated by a strong base
    2. strong base titrated by a strong acid
    3. weak acid titrated by a strong base
    4. weak base titrated by a strong acid
    5. weak base titrated by a weak acid
    Answer

    a. strong acid titrated by a strong base

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

    An initial pH of 13.00, an equivalence point at pH 7.00 and a relatively long, nearly vertical middle section corresponds to a titration curve for _____.

    1. strong acid titrated by a strong base
    2. strong base titrated by strong acid
    3. weak acid titrated by strong acid
    4. weak base titrated by strong acid
    5. weak base titrated by weak acid
    Answer

    b. strong base titrated by strong acid

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

    The pH of a solution prepared by mixing 45mL of 0.183M KOH with a 65mL of 0 .145M HCl is_____?

    Answer

    \[moles\,KOH=0.183*0.045L=0.008235mol \nonumber \]

    \[moles\,HCl=0.145*0.065L=0.009425mol \nonumber \]

    \[moles\,H^{+}=moles\,HCl-moles\,KOH=0.009425mol-0.008235mol=0.00119mol \nonumber \]

    \[\left [ H^{+} \right ]=\frac{0.00119mol}{0.045+0.065}=0.01082M \nonumber \]

    \[pH=-log\left [ H^{+} \right ]=-log(0.01082M)=1.97 \nonumber \]

     

    Titration of Weak Acid by Strong Base

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

    25mL of 0.15 M CH3COOH        Ka = 1.8*10-5

    titrated with 0.15M NaOH

    What is the pH at VB (volume of base) = 0?

    Answer

    \[pH=-log\left [ H^{+} \right ]=-log\sqrt{K_{a}\left [ HA \right ]_{i}}\nonumber \]

    \[pH=-log\sqrt{(1.8*10^{-5})*0.15M}=-log(0.0016431677)=2.78\nonumber \]

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

    25mL of 0.15 M CH3COOH        Ka = 1.8*10-5

    titrated with 0.15M NaOH

    What is the pH at VB = 10mL?

    Answer

    \[pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}\nonumber \]

    \[\left [ A^{-} \right ]=\frac{n_{A^{-}}}{V_{T}}=\frac{0.01L(base\,added)*0.15M(M\,base)}{V_{T}}=\frac{0.0015}{V_{T}}\nonumber \]

    \[\left [ HA \right ]=\frac{n_{HA}}{V_{T}}=\frac{inital\,moles\,of\,acid(0.025L*0.15M)-0.0015(moles\,of\,acid\,used)}{V_{T}}=\frac{0.00225}{V_{T}}\nonumber \]

    \[pH=-log(1.8*10^{-5})+log\frac{\frac{0.0015}{V_{T}}}{\frac{0.00225}{V_{T}}}\nonumber \]

    \[pH=-log(1.8*10^{-5})+log\frac{0.0015}{0.00225}\nonumber \]

    \[pH=4.74472459+(-0.1760912591)\nonumber \]

    \[pH=4.57\nonumber \]

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

    25mL of 0.15 M CH3COOH        Ka = 1.8*10-5

    titrated with 0.15M NaOH

    What is the pH at VB at half equivalence (12.5mL)?

    Answer

    \[pH=pKa\nonumber \]

    \[pH=-log\left ( 1.8*10^{-5} \right )=4.74\nonumber \]

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

    25mL of 0.15 M CH3COOH        Ka = 1.8*10-5

    titrated with 0.15M NaOH

    What is the pH at VB = 15mL?

    Answer

    \[pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}\nonumber \]

    \[\left [ A^{-} \right ]=\frac{0.015L(base\,added)*0.15M(M\,base)}{V_{T}}=\frac{0.00225}{V_{T}}\nonumber \]

    \[\left [ HA \right]=\frac{0.00375(initial\,moles\,of\,acid(0.025L*0.15M))-0.00225(moles\,of\,acid\,used)}{V_{T}}=\frac{0.0015}{V_{T}}\nonumber \]

    \[pH=-log(1.8*10^{-5})+log\frac{\frac{0.00225}{V_{T}}}{\frac{0.0015}{V_{T}}}\nonumber \]

    \[pH=-log(1.8*10^{-5})+log\frac{0.00225}{0.0015}\nonumber \]

    \[pH=4.744727459+(0.1760912591)\nonumber \]

    \[pH=4.92\nonumber \]

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

    25mL of 0.15 M CH3COOH        Ka = 1.8*10-5

    titrated with 0.15M NaOH

    What is the pH at VB at equivalence (25mL)?

    Answer

    \[\left [ OH^{-} \right ]=\sqrt{K_{b}\left [ A^{-} \right ]}\nonumber \]

    \[K_{b}=\frac{10^{-14}}{K_{a}}=\frac{10^{-14}}{1.8*10^{-5}}=5.56*10^{-10}\nonumber \]

    \[\left [ A^{-} \right ]=\frac{0.025L*0.15M}{V_{T}}=\frac{0.00375mol}{0.05L}=0.075M\nonumber \]

    \[pOH=-log\sqrt{K_{b}\left [ A^{-} \right ]}\nonumber \]

    \[pOH=-log\sqrt{5.56*10^{-10}*\left [ 0.075 \right ]}\nonumber \]

    \[pOH=-log(6.458*10^{-6})=5.19\nonumber \]

    \[pH=14-pOH=14-5.19=8.81\nonumber \]

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

    25mL of 0.15 M CH3COOH        Ka = 1.8*10-5

    titrated with 0.15M NaOH

    What is the pH at VB = 30mL (excess base)?

    Answer

    \[Base: 0.03L*0.15M=\left [ 0.0045 \right ]\nonumber \]

    \[Acid: 0.025L*0.15M=\left [ 0.00375 \right ]\nonumber \]

    \[\left [ OH^{-} \right ]=\frac{\left [ 0.0045 \right ]-\left [ 0.00375 \right ]}{0.03+0.025}=0.0136363636M\nonumber \]

    \[pOH=-log(0.0136363636M)=1.87\nonumber \]

    \[pH=14-pOH\nonumber \]

    \[pH=14-1.87=12.13\nonumber \]

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

    What is the molarity of an HOAc solution if 25.5mL of this solution required 37.5mL of 0.175M NaOH to reach the equivalence point?

    Answer

    \[M_{1}V_{1}=M_{2}V_{2}\nonumber \]

    \[M_{2}=\frac{M_{1}V_{1}}{V_{2}}=\frac{0.175M*0.0375L}{0.0255L}=0.257M\nonumber \]

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

    Consider an experiment where 35.0mL of 0.175M HOAc is titrated when 0.25M in NaOH. What is the pH at equivalence point for this titration? The Ka for HOAc is 1.8 * 10-5.

    Answer

    \[M_{1}V_{1}=M_{2}V_{2} V_{2}=\frac{M_{1}V_{1}}{M_{2}}=\frac{0.175M*35.0mL}{0.25M}=24.5mL \nonumber \]

    \[moles\,NaAc=0.175M*0.0350L=0.006125mol \nonumber \]

    \[\left [ NaAc \right ]=\frac{0.006125mol}{0.0350L+.0245L}=0.1039M \nonumber \]

    R Ac-(aq)     + H2O(l) HAc(aq)     + OH-(aq)
    I 0.1039M ---   0 0
    C -x ---   +x +x
    E 0.1039-x ---   x x

    \[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{1.8*10^{-5}}=5.556*10^[-10] \nonumber \]

    \[K_{b}=\frac{\left [ HAc \right ]\left [ OH^{-} \right ]}{\left [ Ac^{-} \right ]}=\frac{x^{2}}{0.1039} \nonumber \]

    \[x=7.5978*10^{-6}M \nonumber \]

    \[pH=14-pOH=14-(-log\left [ OH^{-} \right ])=14-(-log\left [ 7.5978*10^{-6} \right ])=8.88\nonumber \]

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

    50.50mL of 0.116 M HF is titrated with 0.1200 M NaOH. (Ka for HF is 6.8 * 10-4)

    1. How many mL of the base are required to reach the equivalence point?
    2. What is the pH after 50.50ML of base has been added?
    Answer a.

    \[M_{1}V_{1}=M_{2}V_{2}\nonumber \]

    \[V_{2}=\frac{M_{1}V_{1}}{M_{2}}=\frac{0.116M*0.0505L}{0.1200M}=48.82mL\nonumber \]

    Answer b.

    \[\left [ Base \right ]=0.1200M*0.0505L=0.00606mol \nonumber \]

    \[\left [ Acid \right ]=0.116M*0.0505L=0.005858mol \nonumber \]

    \[\left [ OH^{-} \right ]=\frac{0.00606mol-0.005858mol}{0.0505L+0.0505L}=0.002M \nonumber \]

    \[pOH=-log\left [ OH^{-} \right ]=-log(0.002M)=2.700 \nonumber \]

    \[pH=14-pOH=14-2.700=11.300\nonumber \]

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

    65.50mL of 0.161 M HF is titrated where 0.1200 M NaOH. What is the pH at the equivalence point? (Ka for HF is 6.8 * 10-4)

    Answer

    \[M_{1}V_{1}=M_{2}V_{2}\nonumber \]

    \[V_{2}=\frac{M_{1}V_{1}}{M_{2}}=\frac{65.50mL*0.161M}{0.1200M}=87.88mL\nonumber \]

    \[moles,F^{-}=0.06550L*0.161M=0.0105mol\nonumber \]

    \[V_{eq}=65.50mL+87.88mL=153.38mL=0.15338L\nonumber \]

    \[\left [ F^{-} \right ]=\frac{0.0105mol}{0.15338L}=0.0685M\nonumber \]

    R F-(aq)     + H2O(l)   HF(aq)     + OH-(aq)
    I 0.0685M ---   0 0
    C -x ---   +x +x
    E 0.0685-x ---   x x

    \[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{6.8*10^{-4}}=1.47*10^{-11}\nonumber \]

    \[K_{b}=\frac{\left [ HF \right ]\left [ OH^{-} \right ]}{\left [ F^{-} \right ]}=\frac{x^{2}}{0.0685-x}=1.47*10^{-11}\nonumber \]

    \[x=1.00*10^{-6}M\nonumber \]

    \[\left [ H^{+} \right ]=\frac{K_{w}}{\left [ OH^{-} \right ]}=\frac{10^{-14}}{1.00*10^{-6}}=1.00*10^{-8}M\nonumber \]

    \[pH=-log\left [ H^{+} \right ]=-log(1.00*10^{-8}M)=8.002\nonumber \]

     

    Titration of Weak Base with Strong Acid

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

    25mL of 0.15M NH3         Kb = 1.8*10-5

    titrated with 0.15M HCl

    What is the pH at VA (volume of acid) = 0?

    Answer

    \[\left [ OH^{-} \right ]=\sqrt{K_{b}*\left [ NH_{3} \right ]}=\sqrt{1.8*10^{-5}*\left [ 0.15 \right ]}\nonumber\]

    \[pOH=-log\left ( \sqrt{1.8*10^{-5}*\left [ 0.15 \right ]} \right )=2.78\nonumber\]

    \[pH=14-pOH=14-2.78=11.22\nonumber\]

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

    25mL of 0.15M NH3         Kb = 1.8*10-5

    titrated with 0.15M HCl

    What is the pH at VA = 10mL?

    Answer

    \[pOH=pK_{b}+log\frac{base}{salt}\nonumber\]

    \[pOH=-log(1.8*10^{-5})+log\frac{\frac{0.0015}{V_{T}}}{0.00375-\frac{0.0015}{V_{T}}}\nonumber\]

    \[pOH=-log(1.8*10^{-5})+log\frac{0.0015}{0.00225}=4.57\nonumber\]

    \[pH=14-pOH=14-4.57=9.43\nonumber\]

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

    25mL of 0.15M NH3         Kb = 1.8*10-5

    titrated with 0.15M HCl

    What is the pH at VA at half equivalence (12.5mL)?

    Answer

    \[pOH=pK_{b}\nonumber\]

    \[pOH=-log(1.8*10^{-5})=4.74\nonumber\]

    \[pH=14-pOH=14-4.74=9.26\nonumber\]

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

    25mL of 0.15M NH3         Kb = 1.8*10-5

    titrated with 0.15M HCl

    What is the pH at VA = 15mL?

    Answer

    \[pOH=pK_{b}+log\frac{base}{salt}\nonumber\]

    \[pOH=-log(1.8*10^{-5})+log\frac{\frac{0.00225}{V_{T}}}{0.00375-\frac{0.00225}{V_{T}}}\nonumber\]

    \[pOH=-log(1.8*10^{-5})+log\frac{0.00225}{0.0015}=4.92\nonumber\]

    \[pH=14-pOH=14-4.92=9.08\nonumber\]

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

    25mL of 0.15M NH3         Kb = 1.8*10-5

    titrated with 0.15M HCl

    What is the pH at VA = 25mL?

    Answer

    \[\left [ H_{3}O^{+} \right ]=\sqrt{K_{a}*\left [ salt \right ]}\nonumber\]

    \[pH=-log(\sqrt{K_{a}*\left [ salt \right ]})\nonumber\]

    \[K_{a}=\frac{K_{w}}{K_{b}}=\frac{10^{-14}}{1.8*10^{-5}}=5.6*10^{-10}\nonumber\]

    \[\left [ salt \right ]=\frac{0.025L*0.15M}{0.025+0.025}=0.075M\nonumber\]

    \[pH=-log(\sqrt{5.6*10^{-5}*\left [ 0.075 \right ]})=5.18\nonumber\]

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

    25mL of 0.15M NH3         Kb = 1.8*10-5

    titrated with 0.15M HCl

    What is the pH at VA = 30mL (excess acid)?

    Answer

    \[pH=-log\frac{0.030(0.15)-0.025(0.15)}{0.025+0.03}\nonumber\]

    \[pH=-log\frac{0.005(0.15)}{0.025+0.03}=1.87\nonumber\]

     

    17.4: Solubility of Salts

    Textbook: Section 17.4

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

    In which of the following aqueous solutions would you expect AgCl to have the lowest solubility?

    1. pure water
    2. 0.020 M BaCl2
    3. 0.015 M NaCl
    4. 0.020 M AgNO3
    5. 0.020 M KCl
    Answer

    b. 0.020 M BaCl2

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

    Given the following table Ksp values, determine which compound listed has the greatest solubility.

    Compound

    Ksp

    CdCO3

    5.2*10-12

    Cd(OH)2

    2.5*10-14

    AgI

    8.3*10-17

    Fe(OH)3

    4.0*10-38

    ZnCO3

    1.4*10-11

    1. CdCO3
    2. Cd(OH)2
    3. AgI
    4. Fe(OH)3
    5. ZnCO3
    Answer

    b. Cd(OH)2

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

    The solubility of which one of the following will not be affected by the pH of the solution?

    1. Na3PO4
    2. NaF
    3. KNO3
    4. AlCl3
    5. MnS
    Answer

    c. KNO3

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

    Write the expression relating solubility to Ksp for silver sulfide.

    Answer

    \[Ag_{2}S(s)\rightleftharpoons 2Ag^{+}(aq)+S^{2-}(aq)\nonumber\]

    \[K_{sp}=\left [ Ag^{+} \right ]^{2}\left [ S^{2-} \right ]=(2s)^{2}(s)=4s^{3}\nonumber\]

    \[s=(\frac{K_{sp}}{4})^{\frac{1}{3}}\nonumber\]

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

    In which aqueous system is PbI2 least soluble?

    1. H2O
    2. 0.5 M HI
    3. 0.2 M HI
    4. 1.0 M HNO3
    5. 0.8 M KI
    Answer

    e. 0.8 M KI

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

    Of the substances below, _____ will decrease the solubility of Pb(OH)2 in a saturated solution.

    1. NaNO3
    2. H2O2
    3. HNO3
    4. Pb(NO3)2
    5. NaCl
    Answer

    d. Pb(NO3)2

     

    Solubility Calculations

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

    What is the Pb+2 concentration for a saturated solution of PbSO4?  Ksp=6.3x10-7

    Answer

    \[K_{sp}=\left [ Pb^{2+} \right ]\left [ SO_{4}^{2-} \right ]=\left [ x \right ]\left [ x \right ]=x^{2}\nonumber\]

    \[x=\sqrt{K_{sp}}=\sqrt{6.3*10^{-7}}=7.94*10^{-4}M\nonumber\]

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

    What is the Pb+2 concentration if 0.10mol of Na2SO4 is added to 1L of a saturated solution of PbSO4?

    Answer

    \[K_{sp}=\left [ Pb^{2+} \right ]\left [ SO_{4}^{2-} \right ]=\left [ x \right ]\left [ 0.10M \right ]\nonumber\]

    \[x=\frac{K_{sp}}{\left [ 0.10M \right ]}=\frac{6.3*10^{-7}}{\left [ 0.10M \right ]}=6.3*10^{-6}M\nonumber\]

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

    What is the solubility of Ca3(PO4)2?  Ksp=2.0x10-29

    Answer

    \[Ca_{3}\left ( PO_{4}\right )_{2}(aq)\rightarrow 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber\]

    \[K_{sp}=\left [ Ca^{2+} \right ]^{3}\left [ PO_{4}^{3-} \right ]^{2}=\left [ 3x \right ]^{3}\left [ 2x \right ]^{2}=x^{5}(27)(4)\nonumber\]

    \[x^{5}=\frac{K_{sp}}{108} x=\left ( \frac{K_{sp}}{108} \right )^{\frac{1}{5}}\nonumber\]

    \[x=\left ( \frac{2.0*10^{-29}}{108} \right )^{\frac{1}{5}}=7.1*10^{-7}M\nonumber\]

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

    What is the Ca2+ concentration for a saturated solution of Ca3(PO4)2?

    Answer

    \[Ca_{3}\left ( PO_{4}\right )_{2}(aq)\rightarrow 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber\]

    \[K_{sp}=\left [ Ca^{2+} \right ]^{3}\left [ PO_{4}^{3-} \right ]^{2}=\left [ 3x \right ]^{3}\left [ 2x \right ]^{2}=x^{5}(27)(4)\nonumber\]

    \[x^{5}=\frac{K_{sp}}{108} x=\left ( \frac{K_{sp}}{108} \right )^{\frac{1}{5}}\nonumber\]

    \[x=\left ( \frac{2.0*10^{-29}}{108} \right )^{\frac{1}{5}}=7.1*10^{-7}M\nonumber\]

    \[\left [ Ca^{2+} \right ]=3*(7.1*10^{-7}M)=2.1*10^{-6}M\nonumber\]

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

    What is the PO43- concentration for a saturated solution of Ca3(PO4)2?

    Answer

    \[Ca_{3}\left ( PO_{4}\right )_{2}(aq)\rightarrow 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber\]

    \[K_{sp}=\left [ Ca^{2+} \right ]^{3}\left [ PO_{4}^{3-} \right ]^{2}=\left [ 3x \right ]^{3}\left [ 2x \right ]^{2}=x^{5}(27)(4)\nonumber\]

    \[x^{5}=\frac{K_{sp}}{108} x=\left ( \frac{K_{sp}}{108} \right )^{\frac{1}{5}}\nonumber\]

    \[x=\left ( \frac{2.0*10^{-29}}{108} \right )^{\frac{1}{5}}=7.1*10^{-7}M\nonumber\]

    \[\left [ PO_{4}^{3-} \right ]=3*(7.1*10^{-7}M)=1.4*10^{-6}M\nonumber\]

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

    What is the Ksp for MgF2 if F-=0.00236M?

    Answer

    \[MgF_{2}(aq)\rightarrow Mg^{2}(aq)+2F^{-}(aq)\nonumber\]

    \[K_{sp}=\left [ Mg^{2} \right ]\left [ F^{-}( \right ]^{2}=\left [ x \right ]\left [ 2x( \right ]^{2}\nonumber\]

    \[2x=0.00236M\nonumber\]

    \[x=\frac{0.00236M}{2}=0.00118M\nonumber\]

    \[K_{sp}=\left [ 0.00118M \right ]\left [ 0.00236( \right ]^{2}=6.6*10^{-9}\nonumber\]

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

    If the molar solubility of CaF2 at 25°C is 1.25x10-3mol/L, what is the Ksp at this temperature?

    Answer

    \[CaF_{2}(s)\rightleftharpoons Ca^{2+}(aq)+2F^{-}(aq)\nonumber\]

    \[K_{sp}=\left [ Ca^{2+} \right ]\left [ F^{-} \right ]^{2}=\left [ X \right ]\left [ 2X \right ]^{2}\nonumber\]

    \[K_{sp}=\left [ X \right ]4\left [ X \right ]^{2}=4\left [ X \right ]^{3}=4\left [ 1.25*10^{-5} \right ]^{3}=7.81*10^{-9}\nonumber\]

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

    A saturated solution of NaF contains 4.0 g of salt in 100.0ml of water at 15°C, what is the solubility product for NaF?

    Answer

    \[\frac{\frac{4.0g}{42g/mol}}{0.10L}=0.952M\nonumber\]

    \[NaF(s)\rightleftharpoons Na^{+}(aq)+F^{-}(aq)\nonumber\]

    \[K_{sp}=\left [ Na^{+} \right ]\left [ F^{-} \right ]=0.952^{2}=0.91\nonumber\]

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

    The Ksp for AgBr is 5.0x10-13 at 25°C, what is the molar solubility of AgBr?

    Answer

    \[AgBr(s)\rightleftharpoons Ag^{+}(aq)+Br^{-}(aq)\nonumber\]

    \[K_{sp}=\left [ Ag^{+} \right ]\left [ Br^{-} \right ]=5.0*10^{-13}\nonumber\]

    \[x^{2}=5.0*10^{-13}\nonumber\]

    \[x=7.1*10^{-7}M\nonumber\]

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

    Calculate the molar solubility of AgBr in 0.10M NaBr solution?

    Answer
    R AgBr(s) Ag+(aq)     + Br-(aq)
    I ---   0 0.10M
    C ---   x x
    E ---   x .10+x

    \[K_{sp}=\left [ Ag^{+} \right ]\left [ Br^{-} \right ]=5.0*10^{-13}=\left(0.10+x\right)x\nonumber\]

    \[x=5.0*10^{-12}\nonumber\]

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

    Calculate the solubility of Mn(OH)2 in grams per liter when buffered at pH = 9.50? Ksp= 5.0x10-13

    Answer

    \[\left [ OH^{-} \right ]=10^{-\left ( 14-9.5 \right )}=3.16*10^{-5}\nonumber\]

    \[Mn(OH)_{2}(s)\rightleftharpoons Mn^{2+}(aq)+2OH^{-}(aq)\nonumber\]

    \[K_{sp}=5.10^{-13}=\left [ Mn^{2+} \right ]\left [ OH^{-} \right ]^{2}=\left [ Mn^{2+} \right ]\left [ 3.16*10^{-5}\right ]^{2}\nonumber\]

    \[\left [ Mn^{2+} \right ]5.9*10^{-4}M\nonumber\]

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

    Calculate the concentration of Cu2+ at equilibrium when NH3 is added to a 0.010M CuCl2 solution to produce an equilibrium concentration of [NH3] = 0.02M.  Neglect the volume change when the ammonia is added. Kf=5x1012

    Answer

    Kf=5x1012, which is very very large, so assume almost all the copper is converted to the complex, also note, 

    \[K_{f}=\frac{\left [ Cu(NH_{3})_{4}^{2+} \right ]}{\left [ Cu^{2+} \right ]\left [ NH_{3} \right ]^{4}}\nonumber\]

    \[\left [ Cu^{2+} \right ]=\frac{\left [ Cu(NH_{3})_{4}^{2+} \right ]}{K_{f}\left [ NH_{3} \right ]^{4}}\nonumber\]

    \[\left [ Cu^{2+} \right ]=\frac{\left [ 0.01 \right ]}{5*10^{12}\left [ .02 \right ]^{4}}=1.2*10^{-8}\nonumber\]

    \[-=\frac{x\left ( 0.020 \right )^{4}}{0.010}=2*10^{-13}\nonumber\]

    \[x=1.25*10^{-8}M\nonumber\]

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

    What is the molar solubility of MgC2O4? (Ksp for MgC2O4 is 8.6*10-5)

    Answer

    \[MgC_{2}O_{4}\rightleftharpoons Mg^{2+}+C_{2}O_{4}^{2-}\nonumber\]

    \[K_{sp}=\left [ Mg^{2+} \right ]\left [ C_{2}O_{4}^{2-} \right ]\nonumber\]

    \[K_{sp}=(s)*(s)\nonumber\]

    \[s^{2}=K_{sp}\nonumber\]

    \[s=\sqrt{K_{sp}}=\sqrt{8.6*10^{-5}}=9.27*10^{-3}mol/L\nonumber\]

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

    Calculate the concentration (in M) of iodine ions in a saturated solution of lead(II) iodine (Ksp = 1.39*10-8)

    Answer

    \[PbI_{2}(s)\rightleftharpoons Pb^{2+}(aq)+2I^{-}(aq)\nonumber\]

    \[K_{sp}=\left [ Pb^{2+} \right ]\left [ I^{-} \right ]^{2}\nonumber\]

    \[1.39*10^{-8}=x*(2x)^{2}\nonumber\]

    \[1.39*10^{-8}=4x^{3}\nonumber\]

    \[x^{3}=3.475*10^{-9}\nonumber\]

    \[x=(3.475*10^{-9})^{1/3}=1.515*10^{-3}\nonumber\]

    \[\left [ I^{-} \right ]=2*1.515*10^{-3}=3.03*10^{-3}\nonumber\]

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

    Calculate the molar solubility of silver carbonate (Ksp = 6.15*10-12)

    Answer

    \[Ag_{2}CO_{3}(s)\rightleftharpoons 2Ag^{+}(aq)+CO_{3}^{2-}(aq)\nonumber\]

    \[K_{sp}=\left [ Ag^{+} \right ]^{2}\left [ CO_{3}^{2-} \right ]=(2s)^{2}*s\nonumber\]

    \[6.15*10^{-12}=(2s)^{2}*s\nonumber\]

    \[6.15*10^{-12}=4s^{3}\nonumber\]

    \[s^{3}=1.5375*10^{-12}\nonumber\]

    \[s=1.15*10^{-4}\nonumber\]

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

    The concentration (in M) of bromide ions in a saturated solutions of mercury(II) bromide. Ksp = 8.0*10-20, is _____ M.

    Answer

    \[HgBr_{2}(s)\rightleftharpoons Hg^{2+}(aq)+2Br^{-}(aq)\nonumber\]

    \[K_{sp}=\left [ Hg^{2+} \right ]\left [ Br^{-} \right ]^{2}\nonumber\]

    \[8.0*10^{-20}=s*(2s)^{2}\nonumber\]

    \[8.0*10^{-20}=4s^{3}\nonumber\]

    \[s^{3}=2.0*10^{-20}\nonumber\]

    \[s=2.71*10^{-7}\nonumber\]

    \[\left [ Br^{-} \right ]=2s=2*(2.71*10^{-7})=5.4*10^{-7}M\nonumber\]

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

    The solubility of AuCl3 (as Au3+ and Cl-) in water at 298K is 3.3*10-7 M. Ksp for AuCl3 is _____.

    Answer

    \[AuCl_{3}(s)\rightleftharpoons Au^{3+}(aq)+3Cl^{-}(aq)\nonumber\]

    \[{sp}=\left [ Au^{3+} \right ]\left [ Cl^{-} \right ]^{3}\nonumber\]

    \[K_{sp}=\left [ s \right ]\left [ 3s \right ]^{3}\nonumber\]

    \[K_{sp}=\left [ 3.3*10^{-7} \right ]\left [ 3*(3.3*10^{-7}) \right ]^{3}=3.2*10^{-25}\nonumber\]

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

    What is the molar solubility of PbS? (Ksp (PbS) = 8.0*10-28)

    Answer

    \[PbS(s)\rightleftharpoons Pb^{+}(aq)+S^{-}(aq) \nonumber\]

    \[K_{sp}=\left [ Pb^{+} \right ]\left [ S^{-} \right ] \nonumber\]

    \[8.0*10^{-28}=s*s \nonumber\]

    \[s=2.8*10^{-14}\nonumber\]

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

    Calculate the maximum concentration (in M) of sulfide ions in a solution containing 0.181M of lead ions. The Ksp for lead sulfide is 3.4*10-28.

    Answer

    \[PbS(s)\rightleftharpoons Pb^{+}(aq)+S^{-}(aq) \nonumber\]

    \[K_{sp}=\left [ Pb^{+} \right ]\left [ S^{-} \right ] \nonumber\]

    \[3.4*10^{-28}=0.181*x \nonumber\]

    \[x=1.9*10^{-27}M\nonumber\]

     

    General Questions

    Textbook: Section 17

    Exercise \(\PageIndex{a}\)

    What is the pH of 10-9M HCl (a very dilute strong acid)?

    Answer

    pH = 7, because an acid can not have a pH above 7. All the acid disassociated.

    Exercise \(\PageIndex{b}\)

    What is the pH of 6.0M HCl (super strong acid)?

    Answer

    pH = 0, because H3O+ is the strongest acid that can exist in water, and pH cannot be negative.

    Exercise \(\PageIndex{c}\)

    What is the pH of 10-9M NaOH (very dilute strong base)?

    Answer

    pH = 7, because that is the lowest pH a base can have. All the base disassociated.

    Exercise \(\PageIndex{d}\)

    What is the pH of 6.0M NaOH (very concentrated strong base)?

    Answer

    pH = 14, because OH- is the strongest base that can exist in water, and pH cannot be above 14.

    Exercise \(\PageIndex{e}\)

    What is the pH of 10-6M HCl (dilute strong acid)?

    Answer

    \[pH=-log(10^{-6})=6\nonumber\]

    Exercise \(\PageIndex{f}\)

    What is the pH of 6.0M CH3COOH (concentrated weak acid)?

    Answer

    \[pH=-log((1.8*10^{-5})*6)=1.98\nonumber\]

    Exercise \(\PageIndex{g}\)

    Calculate the fluoride ion concentration (in M) in a 1.0L aqueous solution containing 0.40 mol of HF and 0.10 mol of HCl. (Ka for HF = 6.8*10-4).

    Answer
    R HF(aq)

    H+(aq)     + F-(aq)
    I 0.40 mol   0.10 mol 0
    C -x   +x +x
    E 0.40-x   0.10+x x

    \[K_{a}=\frac{\left [ H^{+} \right ]\left [ F^{-} \right ]}{\left [ HF \right ]}=\frac{\left [ 0.10+x \right ]\left [ x \right ]}{\left [ 0.40-x \right ]}=6.8*10^{-4}\nonumber\]

    \[\frac{0.10(x)}{0.40}=6.8*10^{-4}\nonumber\]

    \[x=\frac{0.40}{0.10}*6.8*10^{-4}=2.7*10^{-3}M\nonumber\]

    Exercise \(\PageIndex{h}\)

    Calculate the pH of 1.0L aqueous solution containing 0.30mol of HF and 0.10mol of HCl. (Ka for HF = 6.8*10-4)

    Answer
    R HF(aq)

    H+(aq)     + F-(aq)
    I 0.30 mol   0.10 mol 0
    C -x   +x +x
    E 0.30-x   0.10+x x

    \[K_{a}=\frac{\left [ H^{+} \right ]\left [ F^{-} \right ]}{\left [ HF \right ]}=\frac{\left [ 0.10+x \right ]\left [ x \right ]}{\left [ 0.30-x \right ]}=6.8*10^{-4}\nonumber\]

    \[\frac{0.10(x)}{0.43}=6.8*10^{-4}\nonumber\]

    \[x=\frac{0.40}{0.10}*6.8*10^{-4}=2.7*10^{-3}M\nonumber\]

    \[pH=-log\left [ H^{+} \right ]=-log\left [ 0.10+2.7*10^{-3}M \right ]=0.99 \nonumber \]

    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, rebelford@ualr.edu. 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

     

     

     

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