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1.8 g is dissolved in 100 ml of water. Find milli moles of sulphate ion per litre of the solution? Report the value to nearest integer. Given, $K_{a 1} a n d K a_{2} o f H_{2} S O_{4} a r e 1 \times 10^{3} a n d 4.0 \times 10^{- 2}$ respectively.
Atomic mass $N a = 23, H = 1, S = 32, O = 16$

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answer is 60.

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Detailed Solution

$\underset{(0.15 - x)}{H S O_{4}^{-}} ⇌ \underset{x M}{H^{+}} + \underset{x M}{S O_{4}^{2 -}} 4 \times 10^{- 2} = \frac{x x}{0.15 - x} \Rightarrow x = 6 \times 10^{- 2} M = 60 m i l l i m o l e / L$

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