Question

A beaker with 1.70×102 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 6.80 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. Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

Answer 1

How much will the pH change? The pKa of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

First, let us calculate the acetate/acetic acid ratio from the buffer equations

V = 170 mL pH = 5

total molarit = 0.1 M

pH = pKa+ log(A-/HA)

5 = 4.75 + log(A-/HA)

and we know A- + HA = 0.1*170 = 17

solve

A-/HA = 10^(5.0-4.75) = 1.77827

A-/HA = 1.77827

A- = 1.77827*HA

and A- + HA = 17

1.77827*HA +  HA =17

HA= (17)/(1.77827+1) = 6.1189

then

A- = 1.77827*HA = 1.77827*6.1189= 10.881

Now that we have initial amount of each...

clacualte mmol of A- reacted and mmol of HA increase

mmol of HCl = MV = 6.8*0.310 = 2.108 mmol of HCl

mmol of HA = 6.1189 +2.108 = 8.2269

mmol of A- = 10.881-2.108 = 8.773

substitute in pH equation

pH = pKa+ log(A-/HA)

pH = 4.75+ log(8.773/8.2269)

pH = 4.777