What is [ H +]of a solution having 0.1 M HCN and 0.2 M NaCN ? (Ka. for .HCN =6.2 × 10-10)

Question

What is [ H +]of a solution having 0.1 M HCN and 0.2 M NaCN ? (Ka. for .HCN =6.2 × 10-10) (A)  3.1× 1010  (B)  6.2× 105  (C)  6.2× 10-10  (D)  3.2× 10-9

Tardigrade Admin · Accepted Answer

p H=- log Ka+ log ([N a C N]/H C N) =- log 6.2 × 10-10+ log (0.2/0.1) =- log 6.2+10 log 10+ log 2- log 1 p H=-0.7924+10+0.3010-0 =9.5086 ∵ p H=- log [H+] 0.5086=- log [H+] [H+]=3.225 × 10-9 M

Q. What is $[H^{+}]$ of a solution having $0.1 M H CN$ and $0.2 M N a CN$ ?
$(K_{a}$ for $H CN = 6.2 \times 1 0^{- 10})$

$3.1 \times 10^{10}$

$6.2 \times 10^{5}$

$6.2 \times 10^{- 10}$

$3.2 \times 10^{- 9}$

$p H = - lo g K_{a} + lo g \frac{[ N a CN ]}{H CN}$
$= - lo g 6.2 \times 1 0^{- 10} + lo g \frac{0.2}{0.1}$
$= - lo g 6.2 + 10 lo g 10 + lo g 2 - lo g 1$
$p H = - 0.7924 + 10 + 0.3010 - 0$
$= 9.5086$
$∵ p H = - lo g [H^{+}]$
$0.5086 = - lo g [H^{+}]$
$[H^{+}] = 3.225 \times 1 0^{- 9} M$

Q. What is [H+]of a solution having 0.1MHCN and 0.2MNaCN ? (Ka​ for HCN=6.2×10−10)

3.1×1010

6.2×105

6.2×10−10

3.2×10−9