For the titration of a 25 ml sample of .10 M NH3 by 0.100 HCl. Calculate the PH

Question

For the titration of a 25 ml sample of .10 M NH3 by 0.100 HCl. Calculate the PH of NH3 solution at the following point during titration. 1) Pior to the addition of any HCl 2) After the addition of 12 ml of .10 M HCl. 3) After the point of equivalence point. 4) After the addition of 31 ml of .10 M HCl. Kb=1.8x 10-5. For the titration of a 25 ml sample of .10 M NH3 by 0.100 HCl. Calculate the PH of NH3 solution at the following point during titration. 1) Pior to the addition of any HCl 2) After the addition of 12 ml of .10 M HCl. 3) After the point of equivalence point. 4) After the addition of 31 ml of .10 M HCl. Kb=1.8x 10-5. 1) Pior to the addition of any HCl 2) After the addition of 12 ml of .10 M HCl. 3) After the point of equivalence point. 4) After the addition of 31 ml of .10 M HCl. Kb=1.8x 10-5.

Explanation / Answer

Titration

1) prior to addition of HCl

NH3 + H2O <==> NH4+ + OH-

Kb = 1.8 x 10^-5 = x^2/0.1

x = [OH-] = 1.34 x 10^-3 M

pOH = -log[OH-] = 2.87

pH = 14 - pOH = 11.13

2) after 12 ml of 0.1 M HCl added

[NH3] left = 0.1 M x 13 ml/37 ml = 0.035 M

[NH4+] formed = 0.1 M x 12 ml/37 ml = 0.032 M

pH = 9.25 + log(0.035/0.032) = 9.287

3) at equivalence point

[NH4+] = 0.1 M x 25 ml/50 ml = 0.05 M

NH4+ + H2O <==> NH3 + H3O+

Ka = 1 x 106-14/1.8 x 10^-5 = x^2/0.05

x = [H3O+] = 5.27 x 10^-6 M

pH = -log[H3O+] = 5.28

4) after 31 ml of 0.1 M HCl added

excess [H+] = 0.1 M x 6 ml/56 ml = 0.011 M

pH = -log[H+] = 1.97

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