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Standard XII

Chemistry

pH the Power of H

What is the ...

Question

What is the $p H$ of a $0.05 M$ solution of formic acid? $(K_{a} = 1.8 \times 10^{- 4})$

5.0

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1.8

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2.52

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0.5

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Solution

The correct option is B $2.52$
$H C O O H + H_{2} O ⟶ H C O O^{-} + H_{3} O^{+}$
$0.05 M$ 0 0

at equ. $0.05 - x$ $x$ $x$

$K_{a} = 1.8 \times 10^{- 4}$

$K_{a} = \frac{x^{2}}{0.05 - x} = 1.8 \times 10^{- 4}$

if $x << 0.05$ then $(0.05 - x) \sim 0.05$

thus, $x^{2} = 1.8 \times 10^{- 4} \times 0.05$

$x = \sqrt{1.8 \times 0.05 \times 10^{- 4}}$

$= 0.3 \times 10^{- 2}$

$p H = - log (0.3 \times 10^{- 2}) = 2.52$

Hence, option $C$ is correct.

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What is the conc. of acetic acid that must be added to 0.5M HCOOH solution so that the dissociation of both acids is the same? [An example of Isohydric solutions] $K_{a}$ of acetic acid $1.8 \times 10^{- 5}, K_{a}$ of formic acid = $2.4 \times 10^{- 4}$

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What is the pH of a 0.05 M solution of formic acid? (Ka=1.8×10−4)

The correct option is B 2.52 HCOOH+H2O⟶HCOO−+H3O+ 0.05M 0 0at equ. 0.05−x x xKa=1.8×10−4Ka=x20.05−x=1.8×10−4if x<<0.05 then (0.05−x)∼0.05thus, x2=1.8×10−4×0.05x=√1.8×0.05×10−4 =0.3×10−2pH=−log(0.3×10−2)=2.52Hence, option C is correct.

What is the $p H$ of a $0.05 M$ solution of formic acid? $(K_{a} = 1.8 \times 10^{- 4})$