pH before the For the titration of 20 mL of 01026 M NHs with

pH before the For the titration of 20 mL of 0.1026 M NHs with 0.0947 M HCL, calculate the addition of titrant, at 10 mL prior to the equivalence point, at the equivalence point, and at 10 mL after the equivalence point. k. (NH) 5.7 x 10-10 f the

Solution

2)

pKb of NH3 = 4.76

a) before addition of HCl :

pOH = 1/2 (pKb - log C)

        = 1/2 (4.76 - log 0.1026)

        = 2.87

pH + pOH = 14

pH = 11.13

b) after addtion of 10 mLbefore equivalence point

millimoles of NH3 = 20 x 0.1026 = 2.052

volume At equivalence point :

2.052 = 0.0947 x V

V = 21.67 mL

volume of HCl = 11.67 mL

mmoles of HCl = 11.67 x 0.0947 = 1.105

NH3        +    H+ ----------------> NH4+

2.052              1.105                       0

0.947     0                         1.105

pOH = pKb + log [NH4+/NH3]

pOH = 4.75 + log (1.105 /0.947)

pOH = 4.83

pH = 9.17

c) At equivalence point :

here salt only remains.

salt concentration = 2.052 / 20 + 21.67 = 0.0492 M

pH = 7 - 1/2 (pKb + log C)

      = 7 - 1/2 (4.76 + log 0.0492)

pH = 5.27

d) after addition of 31.67 mL HCl

mmoles of HCl = 31.67 x 0.0947 = 2.999

here strong acid remains.

concentration = 0.01833 M

pH = - log [0.01833]

pH = 1.74