How much 10.0 M HNO3 must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH to 5.35 ?

Answer 1

We need to know the pKa for acetic acid. Let us take that to be 4.74, and proceed.

From the Henderson-Hasselbalch equation, pH = pKa + log [salt]/[acid].

To obtain a pH of 5.35, we can calculate the required ratio of [salt]/[acid] as follows:

5.35 = 4.74 + log [S]/[A]

log [S]/[A] = 0.61

[S]/[A] = 4.07

For the initial buffer, we have a ratio of S/A = 10, i.e 0.1 M salt/0.01 M acid. Upon addition of H⁺ from HNO₃,

we have H⁺ + acetate- ==> HAc thus reducing the concentration of salt, and increasing the concentration of acid.

0.1 - x/0.01 + x = 4.07

S + A = 0.110 since original concentrations are 0.1 M and 0.01 M

S/A = 4.07

S/0.11 - S = 4.07

S = 0.0810 M and A = 0.020 M to obtain desired ratio

To get 0.0810 M acetate, you must add 0.1 -x = 0.081 = 0.019 moles H⁺

(x L)(10.0 mol/L) = 0.019 moles and x = 0.0019 L = 1.9 mls of HNO₃ to be added assuming we don't consider the small change in volume going from 1.00 L to 1.002 L.