What is the percentage hydrolysis of NaCN in N80 solution when the dissociation constant for HCN is 1-3 - 10-9 and Kw - 1-0 - 10-14

The correct option is A 2.48 We know, h= √(K_hC) and h= √(K_wK_a× #160;C) h= √(10^ 14× 801.3 × 10^ 9× 1) #160; h = 2.48% Hence, t ...

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Standard XII

Chemistry

Salt of Weak Acid and Strong Base

What is the p...

Question

What is the percentage hydrolysis of $N a C N$ in $\frac{N}{80}$ solution when the dissociation constant for $H C N$ is $1.3 \times 10^{- 9}$ and $K_{w} = 1.0 \times 10^{- 14}$

2.48

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Solution

The correct option is A $2.48$
We know, $h = \sqrt{\frac{K_{h}}{C}}$ and $h = \sqrt{\frac{K_{w}}{K_{a} \times C}}$
$h = \sqrt{\frac{10^{- 14} \times 80}{1.3 \times 10^{- 9} \times 1}}$ $⟹ h = 2.48 %$

Hence, the percentage hydrolysis of $N a C N$ in $\frac{N}{80}$ solution is $2.48 %$ .

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Similar questions

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What is the percentage hydrolysis of NaCN in N80 solution when the dissociation constant for HCN is 1.3×10−9 and Kw=1.0×10−14

The correct option is A 2.48We know, h=√KhC and h=√KwKa×Ch=√10−14×801.3×10−9×1 ⟹h=2.48%Hence, the percentage hydrolysis of NaCN in N80 solution is 2.48%.

What is the percentage hydrolysis of $N a C N$ in $\frac{N}{80}$ solution when the dissociation constant for $H C N$ is $1.3 \times 10^{- 9}$ and $K_{w} = 1.0 \times 10^{- 14}$

The correct option is A $2.48$
We know, $h = \sqrt{\frac{K_{h}}{C}}$ and $h = \sqrt{\frac{K_{w}}{K_{a} \times C}}$
$h = \sqrt{\frac{10^{- 14} \times 80}{1.3 \times 10^{- 9} \times 1}}$ $⟹ h = 2.48 %$

Hence, the percentage hydrolysis of $N a C N$ in $\frac{N}{80}$ solution is $2.48 %$ .