Question

How many milliliter of 0.1M NaOH should be added to 50.0 ml of 0.1M formic acid HCOOH (Ka=1.8*10^-4), to obtain a buffer with a pH of 4.0?

Answer 1

Answer: The volume of NaOH that must be added is 32.3 mL

Explanation:

Let us assume that volume of NaOH required is 'V' mL

To calculate the millimoles of NaOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mili moles of solute}}{\text{Volume of solution (in mL)}}$

• For NaOH:

Molarity of NaOH solution = 0.1 M

Volume of solution = V mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Mili moles of NaOH}}{V}\\\\\text{Mili moles of NaOH}=0.1V$

• For HCCOH:

Molarity of HCOOH solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Mili moles of HCOOH}}{50}\\\\\text{Mili moles of HCCOH}=5mmol$

The chemical equation for the reaction of formic acid and sodium hydroxide follows:

                       $HCOOH+NaOH\rightarrow HCOONa+H_2O$

Initial:                    5           0.1V

Final:              5 - 0.1 V        -                    0.1V         -

• To calculate the $pK_a$ of acid, we use the equation:

$pK_a=-\log(K_a)$

where,

$K_a$ = acid dissociation constant = $1.8\times 10^{-4}$

Putting values in above equation, we get:

$pK_a=-\log(1.8\time 10^{-4})\\\\pK_a=3.74$

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

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

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

We are given:

$pK_a=3.74$

[HCOONa] = 0.1V

[HCOOH] = 5 - 0.1V

pH = 4.0

Putting values in above equation, we get:

$4.0=3.74+\log(\frac{0.1V}{(5-0.1V)})\\\\V=32.3mL$

Hence, the volume of NaOH that must be added is 32.3 mL