Question

The pka of HCN is 9.3.The pH of a solution
prepared by mixing 2.5 moles of HCN and 0.25
moles of KCN in water and making of the total
volume to 500 ml is
1) 9.3 2) 7.3 3) 10.3
4)8.3
​

Answer 1

Answer:

b............................

Answer 2

The pH of the solution is 8.3

Explanation:

We are given:

Moles of HCN = 2.5 moles

Moles of KCN = 0.25 moles

Volume of solution = 500 mL = 0.5 L     (Conversion factor:  1 L = 1000 mL)

The chemical reaction for HCN and KOH follows the equation:

$HCN+KOH\rightarrow KCN+H_2O$

To calculate the molarity of solution, we use the equation:

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

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

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

$pH=pK_a+\log(\frac{[KCN]}{[HCN]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of hydrogen cyanide = 9.3

$[KCN]=\frac{0.25}{0.5}$

$[HCN]=\frac{2.5}{0.5}$

pH = ?

Putting values in above equation, we get:

$pH=9.3+\log(\frac{0.25/0.5}{2.5/0.5})\\\\pH=8.3$

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