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Molarity

100 ml of N...

Question

$100 m l$ of ${N a}_{2} {C O}_{3}$ solution containing $5.3 g$ of ${N a}_{2} {C O}_{3}$ was exactly neutralised by $200 m l$ of $H_{2} {S O}_{4}$ solution. What is the pH of $H_{2} {S O}_{4}$ ?
( $l o g 5 = 0.6990$ )

A

0.5

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B

1

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C

0.6990

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D

0.3010

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Solution

The correct option is D $0.3010$
$H_{2} S O_{4} + N a_{2} C O_{3} \to N a_{2} S O_{4} + H_{2} O + C O_{2}$

mole of $N a_{2} C O_{3} = \frac{5.3}{106} = 0.05$ mole

$0.05 m o l e$ neutralized by $0.05 m o l H_{2} S O_{4}$

Concentration of $H_{2} S O_{4} = \frac{0.05}{0.2 L} = 0.25$

$p H =$ $- l o g [0.25] = 0.6020$

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Similar questions

1.25 g

{N a}_{2} {C O}_{3}

{N a}_{2} {S O}_{4}

is dissolved in

250 m l

25 m l

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{N a}_{2} {C O}_{3}

in this sample is:

25 g m

{N a}_{2} {C O}_{3}

{N a}_{2} {S O}_{4}

are dissolved in

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{N a}_{2} {C O}_{3}

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{N a}_{2} {C O}_{3}

which can be neutralised by 100 ml of 0.05M

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. 10 ml of this solution required 2.5 ml of 0.1 M

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