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

Any help would be appreciated.

Answer 1

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+ >> 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

Answer 2

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