A buffer that contains 0.35 M of acid, HY and 0.3 M of its conjugate base Y-

Question

A buffer that contains 0.35 M of acid, HY and 0.3 M of its conjugate base Y^-, has a pH of 4.8. What is the pH after 0.018 mol of Ba(OH)₂ are added to 0.82 L of the solution?

The answer is 4.92

Answer 1

First, calculate the pKa of the weak acid.

pH=pKa-log(HA/A-)

4.8=pKa-log(.35/.3)

4.87=pKa

pH=pKa-log(HA/A-) - you know the pKa. You just need to find HA and A-.

If you set up an ICE table, you should be able to calculate the molarity of H+ (using Kw from the amount of OH- present in Ba(OH)2), which should add to the current A- amount. The HA will be this same amount subtracted. After plugging everything in, you should be able to solve for pH.

Let me know if you have questions - it's a bit tough to explain on a message board! Or feel free to set up a call. Buffers are not easy.