Calculate the pH after 0.15 mole of NaOH is added to 1.08 L of a solution that is 0.52 M HC3H5O3 and 1.11 M KC3H5O3... full question below

Question

Calculate the pH after 0.15 mole of NaOH is added to 1.08 L of a solution that is 0.52 M HC₃H₅O₃ and 1.11 M KC₃H₅O₃, and calculate the pH after 0.30 mole of HCl is added to 1.08 L of the same solution of HC₃H₅O₃ and KC₃H₅O₃.

(a)0.15 mole of NaOH (b) 0.30 mole of HCl

Answer 1

A solution of HC₃H₅O₃ and KC₃H₅O₃ constitutes a BUFFER because you have a weak acid HC₃H₅O₃ and the salt of that acid, KC₃H₅O₃. To find the pH after addition of 0.15 moles of NaOH, we can use an ICE table followed by the Henderson Hasselbalch equation, pH = pKa + log [salt]/[acid]

Before setting up the ICE table, we should calculate the moles of each species present.

moles HC₃H₅O₃ = 0.52 mol/L x 1.08 L = 0.5616 moles

moles KC₃H₅O₃ = 1.11 mol/L x 1.08 L = 1.199 moles

Now, for the ICE table using NaOH addition:

HC₃H₅O₃ + NaOH ===> NaC₃H₅O₃ + H2O

0.5616.......0.15.................1.199..................Initial

-0.15.........-0.15..............+0.15..................Change

0.4116........0...................1.349.................Equilibrium

From this ICE table, we can now calculate the final concentrations as follows:

[HC₃H₅O₃] = 0.4116 mol/1.08L = 0.381 M

[C3H5O3^-] = 1.349 mol/1.08L = 1.249 M

Using the HH equation and plugging in our values, we can solve for pH after lookin up the pKa for our acid. The pKa for lactic acid that I found is 3.1. Your value may be different depending on your source.

pH = pKa + log [salt]/[acid]

pH = 3.1 + log (1.249/0.381) = 3.1 + 0.516

pH = 3.62

Follow the identical procedure as above using 0.3 mol HCl and the following ICE table:

C₃H₅O₃^- + H⁺ ===> HC₃H₅O₃

1.199........0.3................0..........Initial

-0.3.........-0.3..............+0.3......Change

0.899.......0...................0.3.......Equilibrium

Convert these equilibrium values to M by dividing by 1.08 L and then use the HH equation to solve for pH