Question

1. A 28.5 mL sample of 0.218 M ethylamine, C₂H₅NH₂, is titrated with 0.340 M nitric acid.

After adding 26.9 mL of nitric acid, the pH is _____.

2. A 24.7 mL sample of 0.292 M methylamine, CH₃NH₂, is titrated with 0.325 M perchloric acid.

At the equivalence point, the pH is _____.

Answer 1

1)

millimoles of ethylamine,= 28.5 x 0.218 = 6.213

millimoles of HNO3 = 26.9 x 0.340 = 9.146

C2H5NH2 + HNO3   ----------------> C2H5NH3+ NO3-

6.213             9.146                               0    

   0    2.933    6.213

here strong acid reamins. so

[HNO3] = 2.933 / (28.5 + 26.9) = 0.0529 M

pH = -log (0.0529)

pH = 1.28

2)

at equivalence point :

millimoles of acid = millimoles of base

0.325 x V = 24.7 x 0.292

V = 22.192 mL

here salt only remians.

[salt] = 24.7 x 0.292 / (24.7 + 22.192) = 0.1538 M

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

     = 7 - 1/2 (3.30 + log 0.1538)

pH = 5.76