Respuesta :
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:
[tex]\text{Molarity of the solution}=\frac{\text{Mili moles of solute}}{\text{Volume of solution (in mL)}}[/tex]
- For NaOH:
Molarity of NaOH solution = 0.1 M
Volume of solution = V mL
Putting values in above equation, we get:
[tex]0.1M=\frac{\text{Mili moles of NaOH}}{V}\\\\\text{Mili moles of NaOH}=0.1V[/tex]
- For HCCOH:
Molarity of HCOOH solution = 0.1 M
Volume of solution = 50 mL
Putting values in above equation, we get:
[tex]0.1M=\frac{\text{Mili moles of HCOOH}}{50}\\\\\text{Mili moles of HCCOH}=5mmol[/tex]
The chemical equation for the reaction of formic acid and sodium hydroxide follows:
[tex]HCOOH+NaOH\rightarrow HCOONa+H_2O[/tex]
Initial: 5 0.1V
Final: 5 - 0.1 V - 0.1V -
- To calculate the [tex]pK_a[/tex] of acid, we use the equation:
[tex]pK_a=-\log(K_a)[/tex]
where,
[tex]K_a[/tex] = acid dissociation constant = [tex]1.8\times 10^{-4}[/tex]
Putting values in above equation, we get:
[tex]pK_a=-\log(1.8\time 10^{-4})\\\\pK_a=3.74[/tex]
- To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[HCOONa]}{[HCOOH]})[/tex]
We are given:
[tex]pK_a=3.74[/tex]
[HCOONa] = 0.1V
[HCOOH] = 5 - 0.1V
pH = 4.0
Putting values in above equation, we get:
[tex]4.0=3.74+\log(\frac{0.1V}{(5-0.1V)})\\\\V=32.3mL[/tex]
Hence, the volume of NaOH that must be added is 32.3 mL
