Question

only 15.61 please

Questions and Exercises 675 OBJECTIVE Calculate the concentrations of the species present in a weak acid solution. 15.61 Write the iCe table and set up the equation needed to solve for the concentration of the hydrogen ion in the fol- lowing solutions. (a) 0.20 M CH COOH (b) 1.50 M HCOOH (c) 0.0055 M HCN (4) 0.075 M HNO, 15.62 Write the iCe table and set up the equation needed to solve for the concentration of the hydrogen ion in the fol- lowing solutions. (a) 1.25 M HOCI (b) 0.80 M HF (c) 0.14 M CH3COOH (d) 0.25 M HCOOH 15.63 A Use the K, values in Table 15.6 to calculate the pH of the following solutions. (a) 0.33 M HNO2 (b) 0.016 M phenol, C H OH (c) 0.25 MHF (d) 0.010 M HCOOH LA Use the K, values in Table 15.6 to calculate the pH of the following solutions. (a) 0.050 MHI (b) 0.85 MHF (c) 0.15 M CH3COOH(d) 0.017 M C H COOH

Answer 1

15.61)

a) C6H5COOH (aq) ----------------------> C6H5COO- (aq) + H+(aq)

      0.20                                                        0                        0 ----------> I

      -x                                                         +x                       +x ------------> C

0.20 -x                                                       x                           x   ----------> E

Ka = [C6H5COO-] [H+] / [C6H5COOH]

Ka = x^2 / 0.20 -x

here x = hydrogen ion concentration

b) HCOOH (aq) ---------------------->         HCOO- (aq) + H+(aq)

      1.50                                                        0                        0 ----------> I

      -x                                                           +x                       +x ------------> C

   1.50 -x                                                       x                           x   ----------> E

Ka = [HCOO-] [H+] / [HCOOH]

Ka = x^2 / 1.50 -x

here x = hydrogen ion concentration

c)

      HCN (aq) ----------------------> CN- (aq) + H+(aq)

      0.0055                                                       0                        0 ----------> I

      -x                                                          +x                       +x ------------> C

     0.0055 -x                                                         x                           x   ----------> E

Ka = [CN-] [H+] / [HCN]

Ka = x^2 / 0.0055 -x

here x = hydrogen ion concentration

d) HNO2 (aq) ---------------------->              NO2- (aq) + H+(aq)

      0.075                                                        0                        0 ----------> I

      -x                                                         +x                       +x ------------> C

0.075 -x                                                       x                           x   ----------> E

Ka = [NO2-] [H+] / [HNO2]

Ka = x^2 / 0.0075 -x

here x = hydrogen ion concentration