Question

alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decimal places.

Answer 1

A 1.0-L buffer solution contains 0.100 mol  HC2H3O2 and 0.100 mol  NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06  

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$  = 0.10 M

The chemical equation for this reaction is :

$\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}$

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is  = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$  

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added  to the original buffer, the concentration of NaOH added will be = $\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}$ = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

                  $\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}$

Initial             0.10               0.035         0.10                  -

Change        -0.035          -0.035       + 0.035              -

Equilibrium    0.065              0              0.135               -

By using  Henderson-Hasselbalch equation:

The pH of this solution = pKa + log $\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}$

The pH of this solution = 4.74 + log $\mathtt{\dfrac{0.135}{0.065}}$

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06    to two decimal places