Question

12. When 0.4 mole of HCl is added to one litre of
solution containing 0.2 M each of HCN and CN-
ion. pH of solution become (K = 1.8 * 10-5)
(1) 3.5
(2) 1
(3) 0.7
(4) 4​

Answer 1

it is acidic buffer solution not affected by strong acid/base addition

Buffet solution has tendency to oppose change in ph

Answer 2

The correct option is 3) 0.7

Explanation:

Molarity is calculated by using the equation:

$Molarity=\frac{Moles}{Volume}$          ...(1)

Given values:

Moles of HCl added = 0.4 moles

Volume of a solution = 1 L

Molarity of HCN = 0.2 M

Molarity of $CN^-$ = 0.2 M

Putting values in equation 1, we get:

Moles of HCN = $(0.2mol/L\times 1L)=0.2 mol$

Moles of $CN^-=(0.2mol/L\times 1L)=0.2 mol$

The chemical equation for the reaction of $CN^-$ with HCl follows:

                   $CN^-+HCl\rightarrow HCN+Cl^-$

I:                   0.2       0.4        0.2

C:                 -0.2     -0.2       +0.2

E:                    -         0.2         0.4

As HCl is a strong acid and HCN is a weak acid. Thus, the pH will be calculated by the hydrogen ions dissociated with the strong acid only.

The chemical equation for the dissociation of HCl follows:

$HCl(aq)\rightarrow H^+(aq)+Cl^-(aq)$

                  0.2 M      0.2 M

To calculate the pH of a solution, we use the equation:

$pH=-\log [H^+]$

Putting values in the above equation, we get:

$pH=-\log (0.2)\\\\pH=0.7$

Hence, the correct option is 3) 0.7

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