Question

A buffer that contains 0.116 M of acid, HY and 0.245 M of its conjugate base Y-, has a pH of 8.83. What is the pH after 0.017 mol of Ba(OH)2 are added to 0.325 L of the solution?

Answer 1

Final pH = 9.99

Explanation

No of mole of HY = (0.116mol/1L)×0.325L=0.0377mol

No of mole of Y- = (0.245mol/1L)×0.325L = 0.0796mol

Ba(OH)2 react with acid HY

Ba(OH)2 + 2HY -------> BaY2 + 2H2O

2 mole of HY react with 1 mole of Ba(OH)2

0.017 mol of Ba(OH)2 react with 0.034mole of HY

after addition

No of mole of HY = 0.0377 - 0.034 = 0.0037mol

No of mole of Y- = 0.0796 + 0.0340 = 0.1136mol

[ Y-] =( 0.1136mol/0.325L)×1L = 0.3495M

[HY] = (0.0037mol/0.325L)×1L = 0.0114M

according to Henderson - Hasselbalch equation

pH= pKa - log (0.3495M/0.0114M) -------> eq1

Henderson - Hasselbalch equation before adding Ba(OH)2

8.83= pKa + log(0.245M/0.116M) --------> eq2

Delate eq2 from1

pH - 8.83 = 0 + log (0.3495/0.0114) - log(0.245/0.116)

= 1.4865 - 0.3247

= 1.16

pH = 8.83 + 1.16 = 9.99