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Answers to Chapter 15 & 16 Study Questions

    1. strong acid + strong base:  steepest at equivalence, low pH at start depending on concentration of acid, equivalence at pH 7, no buffering.
    2. weak acid + strong base:  less steep at equivalence than strong acid, low pH at start depending on \(\mathrm{K_a}\) and concentration of acid, equivalence at basic pH, buffering.
    3. weak base + strong acid:  less steep at equivalence than strong base, high pH at start depending on \(\mathrm{K_b}\) and concentration of base, equivalence at acid pH, buffering.

 

    1. blue 
    2. \(\mathrm{pH =  5}\)
    3. an indicator changes color when \(\mathrm{[HA] = [A^-]}\) and \(\mathrm{[H^+] = K_a}\);
      therefore \(\mathrm{K_a = [H^+] = 10^{-3.5} = 3.2 \times 10^{-4}}\)
    4. weak base + strong acid \(\ce{\rightarrow}\) weak acid;   acid endpoint;  methyl red

 

  1. \(\mathrm{V_A \times M_A =  V_B \times M_B}\);  \(\mathrm{V_A \times 2.00\: M  =  12.5\: mL \times 0.800 \:M}\); \(\mathrm{V_A = 0.500\: mL}\)

(optional pH of solution:)  \(\mathrm{H^+ + NH_3 \rightarrow NH_4^+}\); \(\mathrm{final\: volume = 12.5 + 0.5 = 13.0\: mL}\)

Final \(\mathrm{[NH_4^+]}\):  \(\mathrm{12.5\: mL \times 0.0800\: M = 13.0\: mL \times M_2}\);  \(\mathrm{M_2 = 0.0769\: M\: NH_4^+}\) at end.

So, calculate the pH of 0.0769 M \(\mathrm{NH_4^+}\):

\(\mathrm{K_a =\dfrac{[H^+]\times[NH_3]}{[NH_4^+]}}\); \(\mathrm{5.6\times10^{-10}=\dfrac{x^2}{0.769\:M}}\);  \(\mathrm{x^2 = (5.6 \times 10^{-10})(0.0769) = 4.3 \times 10^{-11}}\);

\(\mathrm{x = (4.3 \times 10^{-11})^{1/2}  = 6.6 \times 10^{-6}\: M}\);  \(\mathrm{pH = - \log (6.6 \times 10^{-6}\: M) = 5.2}\)

(you get the same answer if you used 0.080 M \(\mathrm{NH_4^+}\).)

 

    1. \(\mathrm{K_a(CH_3COOH) = 1.8 \times 10^{-5}}\);  \(\mathrm{[H^+] = 1.0 \times 10^{-5}\: M}\)
      \(\mathrm{\dfrac{[CH_3COO^-]}{[CH_3COOH]}=\dfrac{K_a}{[H^+]}=\dfrac{1.8\times10^{-5}}{1.0\times10^{-5}}=1.8}\)
    2. Mix the solutions in a 1.8: 1 ratio.  For example, mix 180. mL of 0.100 M \(\mathrm{NaCH_3COO}\) with 100 mL 0.100 M \(\mathrm{CH_3COOH}\).
    3. Mix 280 mL 0.100 M \(\mathrm{CH_3COOH}\) and 180. mL 0.100 M \(\ce{NaOH}\).
    4. Another pH 5.00 buffer might be \(\ce{Al(H2O)6^3+}\) or even benzoic acid.

 

    1. limiting reactant problem:  \(\mathrm{0.750\: L \times 0.400\: M\: NaOH = 0.300 \:moles\: OH^-}\)
      \(\mathrm{0.250\: L \times 0.800\: M\: HCl = 0.200\: mol\: H^+}\).
      \(\mathrm{0.300\: moles\: OH^-  + 0.200\: mol\: H^+ \rightarrow  0.200\: mol\: H_2O + 0.100\: mol\: OH^-}\) remain.
      \(\mathrm{Total\: volume = 250\: mL + 750\: mL = 1.00\: liter}\).
      \(\mathrm{[OH^-]=\dfrac{0.100\:mol}{1\:liter}=1.00\:M}\); \(\mathrm{pOH = 1.0}\); \(\mathrm{pH = 13.0}\)
       
    2. \(\mathrm{pH=pK_a+\log\dfrac{[HCO_3^-]}{[H_2CO_3]}}\);     \(\mathrm{pK_a=-\log(4.4\times10^{-7})=6.36}\)
      \(\mathrm{pH=6.36+\log\left( \dfrac{1}{2} \right)= 6.36 - 0.30  = 6.06}\)

 

  1. \(\mathrm{MgCl_­2 \rightleftarrows Mg^{2+}(aq)  +  2  Cl^-(aq)}\);    \(\mathrm{K_{sp} = [Mg^{2+}] \times [Cl^-]^2}\)

\(\mathrm{[MgCl_2]=\dfrac{8.0\:g\:MgCl_2}{108\:g\:solution}\times\dfrac{1\:g\:solution}{1\:mL\:solution}\times\dfrac{1000\:mL}{1\:L}\times\dfrac{1\:mol\:MgCl_2}{95.2\:g\:MgCl_2}=0.778\:M}\)

0.778 M \(\ce{MgCl­2}\):  \(\mathrm{[Mg^{2+}] = 0.778\: M}\);  \(\mathrm{[Cl^-] = 2(0.778\: M) = 1.56\: M}\)

\(\mathrm{K_{sp} = [Mg^{2+}] \times [Cl^-]^2 = (0.778) \times (1.56)^2 = 1.88}\)

 

  1. \(\mathrm{CuCrO_­4 \rightleftarrows Cu^{2+}(aq)  +  CrO_4^{2-}(aq)}\);    \(\mathrm{K_{sp} =  [Cu^{2+}] \times [CrO_4^{2-}]  =  3.6 \times 10^{-6}}\); 

\(\mathrm{x = [CuCrO­_4] = [Cu^{2+}] = [CrO_4^{2-}]}\);  \(\mathrm{K_{sp} = 3.6 \times 10^{-6} = x^2}\)

\(\mathrm{x = (3.6 \times 10^{-6})^{1/2}  = 1.9 \times 10^{-3}\: M}\)

 

  1. \(\mathrm{2 AgNO_3(aq) + Na_2CO_3(aq) \rightarrow 2 NaNO_3(aq) + Ag_2CO_3(s)}\)

\(\mathrm{Ag_2CO_3(s) \rightleftarrows 2 Ag^+(aq) + CO_3^{2-}(aq)}\);  \(\mathrm{K_{sp} = [Ag^+]^2 \times [CO_3^{2-}] = 8.1 \times 10^{-12}}\)

\(\mathrm{x = [Ag^+]}\);  \(\mathrm{[CO_3^{2-}] = 0.020\: M}\);  \(\mathrm{K_{sp} = x^2(0.020) = 8.1 \times 10^{-12}}\)

\(\mathrm{x^2=\dfrac{8.1\times10^{-12}}{0.020}= 4.1 \times 10^{-10}}\);  \(\mathrm{x  =  (4.1 \times 10^{-10})^{1/2}  = 2.0 \times 10^{-5}\: M}\)

 

  1. \(\mathrm{Pb(NO_3)_2(aq) + 2 NaBr(aq) \rightarrow 2 NaNO_3(aq) + PbBr_2(s)}\).

\(\mathrm{PbBr_2(s) \rightleftarrows Pb^{2+}(aq) + 2 Br^-(aq)}\);    \(\mathrm{K_{sp} = [Pb^{2+}] \times [Br^-]^2}\);   \(\mathrm{K_{sp} = 4.6 \times 10^{-6}}\)

When solutions are mixed in a 1:1 ratio, each is diluted by a factor of 2.

\(\mathrm{[Pb^{2+}] = \dfrac{0.0100\: M}{2} = 0.00500 \:M}\);  \(\mathrm{[Br^-] = \dfrac{0.0200\: M}{2} = 0.0100\: M}\)

\(\mathrm{Q = [Pb^{2+}] \times [Br^-]^2 = (0.00500) \times (0.0100)^2  =  5.0 \times 10^{-7}}\)

\(\mathrm{5.0 \times 10^{-7} < 4.6 \times 10^{-6}}\);  \(\mathrm{Q < K_{sp}}\);  therefore, no precipitate forms

 

    1. strong acid reacts with the weak acid in the buffer:

\(\mathrm{H^+(aq) + HCO_3^-(aq) \rightarrow H_2CO_3(aq)}\)

  1. \(\mathrm{Ag_3PO_4(s) \rightleftarrows 3 Ag^+(aq) + PO_4^{3-}(aq)}\);  \(\mathrm{K_{sp} = [Ag^+]^3 \times [PO_4^{2-}]}\)

 

  1. \(\mathrm{molar\: mass = \dfrac{mass}{moles}}\);

\(\textrm{moles acid  =  moles base: }\mathrm{17.5\:mL\times\dfrac{0.268\:mol\:NaOH}{1000\:mL} = 0.00469\: moles\: acid}\)

\(\mathrm{molar\:mass = \dfrac{0.422\:g}{0.00469\:moles}= 90.0\: g/mol}\)