Question

A buffer containing 0.7269 M of acid, HA, and 0.1437 M of its conjugate base, A−, has a pH of 2.89. What is the pH after 0.0014 mol NaOH is added to 0.5000 L of this solution?

I got 4.28 and that was wrong

Answer 1

For a buffer solution pH =pKa + log base/acid

Here put values

        2.89 = pKa + log ( .1437/.7269)

     2.89 = pKa - 0.7040

So pKa = 2.89 +   0.7040       = 3.594

In second case since .0014 mol of NaOH are added to 0.5000L it means adding 0.0028 mol per litre

so molar conc of base = 0.1437 + 0.0028 = 0.1465 M

pH = pKa + log base/acid = 3.594 + log 0.1465/0.7269

      = 3.594 + (-0.6957) = 2.8983