Calculate the pH of 0-2 M NaOCl solution-KaHOCl - 3-5 - 10-8

NaOCl is a salt of weak acid HOCl and strong base NaOH pKa= log(K_a) pK_a= log(3.5×10^ 8) pK_a=(8 0.54)=7.46 pH=7+12(pK_a+logC) pH=7+12(7.46+log(0.2)) pH=7+12(7 ...

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Standard XII

Chemistry

Salt of Weak Acid and Strong Base

Calculate the...

Question

Calculate the $p H$ of 0.2 M $N a O C l$ solution.
$K_{a} (H O C l) = 3.5 \times 10^{- 8}$

10.01

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12.53

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8.63

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9.53

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Solution

The correct option is A $10.01$
$N a O C l$ is a salt of weak acid $H O C l$ and strong base $N a O H$
$p K a = - l o g (K_{a}) p K_{a} = - l o g (3.5 \times 10^{- 8}) p K_{a} = (8 - 0.54) = 7.46 p H = 7 + \frac{1}{2} (p K_{a} + l o g C) p H = 7 + \frac{1}{2} (7.46 + l o g (0.2)) p H = 7 + \frac{1}{2} (7.46 - 0.7) p H = 10.38$

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Salt of Weak Acid and Strong Base

Standard XII Chemistry

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Calculate the pH of 0.2 M NaOCl solution. Ka(HOCl)=3.5×10−8

The correct option is A 10.01NaOCl is a salt of weak acid HOCl and strong base NaOH pKa=−log(Ka)pKa=−log(3.5×10−8)pKa=(8−0.54)=7.46pH=7+12(pKa+logC)pH=7+12(7.46+log(0.2))pH=7+12(7.46−0.7)pH=10.38

Calculate the $p H$ of 0.2 M $N a O C l$ solution.
$K_{a} (H O C l) = 3.5 \times 10^{- 8}$