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Answers to More Chapters 15 & 16 Study Questions

 

1.  \(\mathrm{\dfrac{K_a}{[H^+]}=\dfrac{[C_3H_5O_2^-]}{[HC_3H_5O_2]}=\dfrac{1.3\times10^{-5}}{1.0\times10^{-5}}=1.3}\)

 

2.  \(\mathrm{pH=pK_a+\log\dfrac{[H_2BO_3^-]}{[H_3BO_3]}}\); \(\mathrm{pK_a=-\log(5.8\times10^{-10})=9.24}\); \(\mathrm{pH=9.24+\log\dfrac{1.5}{1}}\)

=   9.24  +  0.18  = 9.42  

 

3.  Na2SO4(aq)  +  Sr(NO3) 2(aq)  →   2 NaNO3(aq)  +  SrSO4(s)

     SrSO4(s)  ⇌  Sr2+(aq)  +  SO42-(aq)         Ksp  =   [Sr2+] x [SO42-]          Ksp = 3.2 x 10-7

x  =  [Na2SO4]  =  [SO42-];  [Sr2+]  =  0.10 M

3.2 x 10-7  =  (0.10 M)(x);  x  =  (3.2 x 10-7)/0.10 M  = 3.2 x 10-6

 

4.  Pb(IO3)2(s)  ⇌ Pb2+(aq)  +  2  IO3-(aq);    Ksp  =   [Pb2+] x [IO3-]2

      [Pb2+] = [Pb(IO3)2] = 2.6 x 10-11 M;   \(\mathrm{[IO_3^-] = [KIO_3] + [Pb(IO_3)_2] \approx  [KIO_3] = 0.10\: M}\)

      Ksp  =   [Pb2+] x [IO3-]2  =  (2.6 x 10-11 M)(0.10 M)2  = 2.6 x 10-13 

 

5.  PbI2(s)  ⇌ Pb2+(aq)  +  2  I-(aq);    Ksp  =   [Pb2+] x [I-]2;   Ksp = 1.4 x 10-8

x  =  [PbI2]  =  [Pb2+];  \(\mathrm{[I^-] = 0.010\: M + 2 [Pb^{2+}] \approx 0.010\: M}\)

     Ksp = [Pb2+] x [I-]2;  1.4 x 10-8 = x [0.010 M]2; x = (1.4 x 10-8)/(1 x 10-4) = 1.4 x 10-4 M

 

6.  PbCl2(s)  ⇌ Pb2+(aq)  +  2  Cl-(aq);    Ksp  =   [Pb2+] x [Cl-]2;   Ksp = 1.7 x 10-5

      x  =  [PbCl2]  =  [Pb2+];  [Cl-]  =  2x;   1.7 x 10-5  =  x (2x)2  =  4 x3

      x3  =  (1.7 x 10-5)/4  =  4.2 x 10-6x  =  (4.2 x 10-6)1/3  = 0.016 M

 

7.  Pb(NO3)2(aq)  +  2NaCl(aq)  →   2 NaNO3(aq)  + PbCl2(s)

     PbCl2(s)  ⇌ Pb2+(aq)  +  2  Cl-(aq);   Ksp  =   [Pb2+] x [Cl-]2;   Ksp = 1.7 x 10-5

     x  =  [Pb(NO3)2]  =  [Pb2+];  [Cl-]  =  [NaCl]  =  0.010 M

     Ksp  =   [Pb2+] x [Cl-]2;   1.7 x 10-5  =  x (0.010)2x  =  (1.7 x 10-5)/( 1 x 10-4)  = 0.17 M

 

8. 

  1. 0.10 M NaOH;  pH  = 13  →   yellow
  2. 0.10 M NaOH;  pH  = 13  →   purple
  3. cresol red is orange when pH = pKa;  pH = 1
  4. yellow in methyl yellow:  pH > 4;  yellow in cresol purple:  pH < 7;  so,  4 < pH < 7

 

9. 

  1. \(\mathrm{moles\: base = moles\: acid = 28.0\: mL \times\dfrac{0.150\:moles\:HCl}{1000\:mL} = 0.00420\: moles}\)
  2. \(\mathrm{molar\: mass =\dfrac{mass}{moles}=\dfrac{0.290\:g}{0.00420\:moles} = 69.0\: g/mole}\)

 

10. 

  1. moles base =  moles acid: VB  x  MB  =  VA  x  MA

2.50 mL  x  3.00 M  =  750 mL  x  MA;  \(\mathrm{M_A = \dfrac{2.50\times3.00}{750}=0.0100\: M\:HCl}\) 

  1. pH  =  - log[H+]  =  - log (1.00 x 10-2 M);  pH  = 2.0