Question

A 500 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 5 mL of 1.00 M KOH. What is the pH following this addition?

Answer 1

Answer: The pH of the solution after addition of KOH is 3.84

Explanation:

We are given:

pH of buffer = 3.75

$pK_a$ of formic acid = 3.75

Using Henderson-Hasselbalch equation for formate buffer:

$pH=pK_a+\log(\frac{[HCOO-]}{[HCOOH]})$

Putting values in above equation, we get:

$3.75=3.75+\log(\frac{[HCOO-]}{[HCOOH]})\\\\\log(\frac{[HCOO-]}{[HCOOH]})=0\\\\\frac{[HCOO-]}{[HCOOH]}=1$

$[HCOO-]=[HCOOH]$

We are given:

Concentration of formate buffer = 0.100 M

$[HCOO-]+[HCOOH]=0.1$

$[HCOO-]=[HCOOH]=0.05M$

As, the volume of buffer is the same. So, the concentration is taken as number of moles of formate ions as well as formic acid

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

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

Molarity of KOH = 1.00 M

Volume of solution = 5 mL

Putting values in above equation, we get:

$1.00M=\frac{\text{Moles of KOH}\times 1000}{5mL}\\\\\text{Moles of KOH}=0.005mol$

The chemical reaction for formic acid and KOH follows the equation:

                  $HCOOH+KOH\rightarrow HCOO^-+H_2O$

Initial:       0.05    0.005               0.05

Final:         0.045          -                0.055          

Volume of solution = 500 + 5 = 505 mL = 0.505 L    (Conversion factor:  1 L = 1000 mL)

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

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

$pH=pK_a+\log(\frac{[HCOO^-]}{[HCOOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of formic acid = 3.75

$[HCOO^-]=\frac{0.055}{0.505}$

$[HCOOH]=\frac{0.045}{0.505}$

pH = ?

Putting values in above equation, we get:

$pH=3.75+\log(\frac{0.055/0.505}{0.045/0.505})\\\\pH=3.84$

Hence, the pH of the solution after addition of KOH is 3.84