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

Step-by-step 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* 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(([salt])/([acid]))`

`pH=pK_a+\log(([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((0.1V)/((5-0.1V)))\\\\V=32.3mL`

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