Question

Calculate the pH of adding 20 mL of 1 M NaOH solution to 100 mL of a 1 M acetic acid (CH3COOH) solution and 880 mL of distilled water. Assume that the Ka of acetic acid is 1.8 * 10-5

Answer 1

Answer: The pH of the solution is 4.14

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

• For NaOH:

Molarity of NaOH = 1 M

Volume of solution = 20 mL

Putting values in above equation, we get:

$1M=\frac{\text{Moles of NaOH}\times 1000}{20mL}\\\\\text{Moles of NaOH}=\frac{1\times 20}{1000}=0.02mol$

• For acetic acid:

Molarity of acetic acid solution = 0.26 M

Volume of solution = 100 mL

Putting values in above equation, we get:

$1M=\frac{\text{Moles of acetic acid}\times 1000}{100mL}\\\\\text{Moles of acetic acid}=\frac{1\times 100}{1000}=0.1mol$

The chemical reaction for NaOH and acetic acid follows the equation:

                   $CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

Initial:           0.100            0.020

Final:            0.080               -              0.020            

Volume of solution = 20 + 100 = 120 mL = 0.120 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=\frac{0.020}{0.120}$

$[CH_3COOH]=\frac{0.080}{0.120}$

pH = ?

Putting values in above equation, we get:

$pH=4.74+\log(\frac{0.020/0.120}{0.080/0.120})\\\\pH=4.14$

Hence, the pH of the solution is 4.14