A beaker with 2.00×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.00 mL of a 0.310 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
millimoles of acid + conjugate base = 2.00 x 10^2 x 0.1 =20 ------------->1
pH = pKa + log [conjugate base / acid ]
5.00 = 4.74 + log [conjugate base / acid ]
1.82 = conjugate base / acid
conjugate base = 1.82 acid -------------> 2
from 1 and 2
1.82 acid + acid = 20
2.82 acid = 19
acid = 7.09
conjugate base = 12.9
millimoles of strong acid = C = 8 x 0.310 = 2.48
on additon of C millimoles of acid to buffer
pH = pKa + log [conjugate base - C / acid + C]
pH = 4.74 + log (12.9 - 2.48 / 7.09+ 2.48)
pH = 4.777
pH chnage = 4.777 - 5.000
= -0.223
pH chnage = - 0.223
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