Find the pH of an equimolar solution of HCN and NH4OH-Given- KaHCN- 6 - 10-10 KbNH4OH-1-8 - 10-5

PK_a= log(K_a) pK_a= log(6× 10^ 10)=(10 0.78) pK_a=9.22 pK_b= log(K_b) pK_b= log(1.8× 10^ 5)=(5 0.26) pK_b=4.74 For a salt of weak acid and weak base nbsp; ...

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Standard XII

Chemistry

Salt of Weak Acid and Weak Base

Find the pH o...

Question

Find the $p H$ of an equimolar solution of $H C N$ and $N H_{4} O H$ .
Given: $K_{a} (H C N) = 6 \times 10^{- 10}$ $K_{b} (N H_{4} O H) = 1.8 \times 10^{- 5}$

8.45

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4.74

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12.5

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9.24

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Solution

The correct option is D 9.24
$p K_{a} = - log (K_{a}) p K_{a} = - log (6 \times 10^{- 10}) = (10 - 0.78) p K_{a} = 9.22$

$p K_{b} = - log (K_{b}) p K_{b} = - log (1.8 \times 10^{- 5}) = (5 - 0.26) p K_{b} = 4.74$

For a salt of weak acid and weak base pH is given as:
$p H = 7 + \frac{1}{2} (p K_{a} - p K_{b}) p H = 7 + \frac{1}{2} (9.22 - 4.74) p H = 7 + \frac{4.48}{2} p H = 9.24$

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Salt of Weak Acid and Weak Base

Standard XII Chemistry

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Find the pH of an equimolar solution of HCN and NH4OH. Given: Ka(HCN)=6×10−10 Kb(NH4OH)=1.8×10−5

The correct option is D 9.24pKa=−log(Ka)pKa=−log(6×10−10)=(10−0.78)pKa=9.22 pKb=−log(Kb)pKb=−log(1.8×10−5)=(5−0.26)pKb=4.74 For a salt of weak acid and weak base pH is given as: pH=7+12(pKa−pKb)pH=7+12(9.22−4.74)pH=7+4.482pH=9.24

Find the $p H$ of an equimolar solution of $H C N$ and $N H_{4} O H$ .
Given: $K_{a} (H C N) = 6 \times 10^{- 10}$ $K_{b} (N H_{4} O H) = 1.8 \times 10^{- 5}$