Ionisation constant of each HA weak acid and BOH weak base are 3-0- 10-7 each at 298 K- The percentage degree of hydrolysis of BA at the dilution of 100 L is-

The correct option is C 25For a salt of weak acid and weak base, BA , #160; h1 h = √(K_h)=√(K_ωK_aK_b) . K_a = K_b = 3 × 10^ 7 and K_w = 1 ...

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Byju's Answer

Standard XII

Chemistry

Dissolution of Solids and Gases in Liquid

Ionisation co...

Question

Ionisation constant of each $H A$ (weak acid) and $B O H$ (weak base) are $3.0 \times 10^{- 7}$ each at 298 K. The percentage degree of hydrolysis of $B A$ at the dilution of 100 L is:

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Solution

The correct option is C 25
For a salt of weak acid and weak base, $B A$ ,

$\frac{h}{1 - h} = \sqrt{K_{h}} = \sqrt{\frac{K_{ω}}{K_{a} K_{b}}}$ .

$K_{a} = K_{b} = 3 \times 10^{- 7}$ and $K_{w} = 10^{- 14}$
Substitute the values in the above expression,
$\frac{h}{1 - h} = \sqrt{K_{h}} = \sqrt{\frac{10^{- 14}}{(3 \times 10^{- 7})^{2}}} = \frac{1}{3}$
Hence, $h = 0.25$ .

The percentage degree of hydrolysis of BA at the dilution of 100 L is 25%

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Ionisation constant of each HA (weak acid) and BOH (weak base) are 3.0×10−7 each at 298 K. The percentage degree of hydrolysis of BA at the dilution of 100 L is:

The correct option is C 25For a salt of weak acid and weak base, BA, h1−h=√Kh=√KωKaKb.Ka=Kb=3×10−7 and Kw=10−14Substitute the values in the above expression,h1−h=√Kh=√10−14(3×10−7)2=13Hence, h=0.25.The percentage degree of hydrolysis of BABA at the dilution of 100 L is 25%

Ionisation constant of each $H A$ (weak acid) and $B O H$ (weak base) are $3.0 \times 10^{- 7}$ each at 298 K. The percentage degree of hydrolysis of $B A$ at the dilution of 100 L is: