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

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Solution

Given: $K_{a} = 1.3 \times 10^{- 9}$ $N = \frac{N}{80}$
$K_{w} = 1 \times 10^{- 14}$
Degree of hydrolysis, $K_{h} = \frac{K_{w}}{K_{a}} \Rightarrow \frac{1 \times 10^{- 14}}{1.3 \times 10^{- 9}} = 0.77 \times 10^{- 5}$
$h = \sqrt{\frac{K h}{c o n c e n t r a t i o n}}$
$= \sqrt{\frac{0.77 \times 10^{- 5}}{1 / 80}}$
$= 24.8 \times 10^{- 3}$
$∴$ Percentage hydrolysis= $24.8 \times 10^{- 3} \times 100 = 2.48$ %

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What is the percentage hydrolysis of NaCN in N/80 solution when the dissociation constant for HCN is 1.3 × 10−9 and Kw = 1.0 ×10−14?

Given: Ka=1.3×10−9 N=N80 Kw=1×10−14Degree of hydrolysis, Kh=KwKa⇒1×10−141.3×10−9=0.77×10−5h=√Khconcentration =√0.77×10−51/80 =24.8×10−3∴ Percentage hydrolysis= 24.8×10−3×100=2.48%

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