A weak acid HX has the dissociation constant 1times - 10 - -5 -M- It forms a salt NaX in reaction with alkali- The degree of hydrolysis of 0-1M solution NaX is-0-0001-0-15-0-01-0-1-

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Toppr Admin · Accepted Answer

X-x2212-H2O-x21CC-HX-OH-x2212-Kh-10-x2212-1410-x2212-5So- degree of hydrolysisx-x221A-KhC-x221A-10-x2212-910-x2212-1-10-x2212-4- degree of hydrolysis -10-x2212-4-xD7-100-0-01

A weak acid $H X$ has the dissociation constant $1 \times 10^{- 5} M$ , It forms a salt $N a X$ in reaction with alkali. The degree of hydrolysis of $0.1 M$ solution if $N a X$ is:
$0.0001$ %
$0.15$ %
$0.01$ %
$0.1$ %

$X^{-} + H_{2} O ⇌ H X + {O H}^{-}$

$K_{h} = \frac{10^{- 14}}{10^{- 5}}$

So, degree of hydrolysis
$x = \sqrt{\frac{K_{h}}{C}} = \sqrt{\frac{10^{- 9}}{10^{- 1}}} = 10^{- 4}$

% degree of hydrolysis $= 10^{- 4} \times 100 = 0.01$

A weak acid HX has the dissociation constant 1×10−5M, It forms a salt NaX in reaction with alkali. The degree of hydrolysis of 0.1M solution if NaX is:0.0001%0.15%0.01%0.1%

X−+H2O⇌HX+OH−Kh=10−1410−5So, degree of hydrolysisx=√KhC=√10−910−1=10−4% degree of hydrolysis =10−4×100=0.01

A weak acid $H X$ has the dissociation constant $1 \times 10^{- 5} M$ , It forms a salt $N a X$ in reaction with alkali. The degree of hydrolysis of $0.1 M$ solution if $N a X$ is:
$0.0001$ %
$0.15$ %
$0.01$ %
$0.1$ %

$X^{-} + H_{2} O ⇌ H X + {O H}^{-}$

$K_{h} = \frac{10^{- 14}}{10^{- 5}}$

So, degree of hydrolysis
$x = \sqrt{\frac{K_{h}}{C}} = \sqrt{\frac{10^{- 9}}{10^{- 1}}} = 10^{- 4}$

% degree of hydrolysis $= 10^{- 4} \times 100 = 0.01$