Adding a strong acid to a buffer

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 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.



Any help would be appreciated.Adding a strong acid to a buffer
the pH of the buffer is dependent only on the formal ratio of the base to the acid form by the relationshippH = pKa + log [base]/[acid]

Since the total molarity is 0.1 and you have 175 mL of the buffer then the total mmoles of both forms together is

175 x 0.1M = 17.5 mmoles so the ratio will be the base to the acid and the two forms must add up to 17.5 mmoles



so pH = 4.76 + log base/17.5 - base

5- 4.76 = log base /17.5 - base = 1.74 = base /17.5 -base



base = 11.11 mmoles therefore acid = 17.5 -11.11 = 6.39



check



pH = 4.76 + log {11.11/6.39 } = 5.0002



so adding 7.3 mL of 0.34 M HCl will add 7.3 X .34 = 2.482 mmoles of H+ to the buffer..This will increase the acid form of the buffer by reacting with 2.482 mmoles of the base form to produce 2.482 mmoles of the acid form..sooo the total mmoles will NOT change only the ratio will and thus the pH...





soo pH = 4.76 + log{ (11.11-2.482 )}/ 6.39 + 2.482 = l4.76 + log 8.628/8.872 = 4.748







Adding a strong acid to a buffer
pH = pKa + log [CH3COO-] / [CH3COOH]



5.00 = 4.74 + log [CH3COO-]/ [CH3COOH]



[CH3COO-] / [CH3COOH] = 10^0.26 = 1.82



[CH3COOH] + [CH3COO-] = 0.1



we must solve this system :

[CH3COO-] = 0.0645 M and [CH3COOH] = 0.0355 M



moles CH3COO- = 0.0645 x 0.175 = 0.0113

moles CH3COOH = 0.0355 x 0.175 = 0.00621



CH3COO- + H+ %26gt;%26gt; CH3COOH

moles HCl = 0.340 x 0.00730 L = 0.00248



moles CH3COO- = 0.0113 - 0.00248 = 0.00882

moles CH3COOH = 0.00621 + 0.00248 = 0.00869



total volume = 0.1823 L



[CH3COO- ] = 0.00882/ 0.1823 L = 0.0484 M

[CH3COOH] = 0.00869 / 0.1823 = 0.0477 M



pH = 4.74 + log 0.0484/ 0.0477 = 4.75