Adding strong acid to a buffer

Question

A beaker with 175 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 7.30 mL of a 0.340M solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.

Answer 1

5.00 = 4.76 + log [CH3COO-] / [CH3COOH]

[CH3COO-] / [CH3COOH] = 10^0.24 = 1.74

[CH3COO-] + [CH3COOH] = 0.1

[CH3COO-] / [CH3COOH] = 1.74

solving this system we get [CH3COO-] = 0.0635 M and [CH3COOH] = 0.0365 M

Moles CH3COO- = 0.0635 x 0.175 = 0.0111

moles CH3COOH = 0.0365 x 0.175 = 0.00639

Now you forgot to say if the solution is an acid or a base so I can not answre at your question but I thing that this can help you