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Osmotic Pressure

300 mL of H...

Question

$300$ mL of $H_{2} S O_{4}$ solution of certain molarity was mixed with $200$ mL of a $B a (O H)_{2}$ solution of same molarity. The pH of the mixture solution is $1$ . The normality of $H_{2} S O_{4}$ and $B a (O H)_{2}$ is $x \times 10^{- 1} N$ , where $x$ is:

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Solution

The pH of the mixture is $1$ .
Hence, the hydrogen ion concentration and the normality of the resulting solution is $0.1 N$ .

Let $n$ be the normality of sulfuric acid and barium hydroxide solution.

The normality of the resulting solution is given by the expression $\frac{N_{a} V_{a} - N_{b} V_{b}}{V_{a} + V_{b}}$ .

Substituting values in the above expression, we get-

$0.1 = \frac{n \times 300 - n \times 200}{300 + 200}$

Thus, $n = 0.5 N = 5 \times 10^{- 1} N$ .

Hence, the value of $x$ is $5$ .

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300 mL of H2SO4 solution of certain molarity was mixed with 200 mL of a Ba(OH)2 solution of same molarity. The pH of the mixture solution is 1. The normality of H2SO4 and Ba(OH)2 is x×10−1N, where x is:

$300$ mL of $H_{2} S O_{4}$ solution of certain molarity was mixed with $200$ mL of a $B a (O H)_{2}$ solution of same molarity. The pH of the mixture solution is $1$ . The normality of $H_{2} S O_{4}$ and $B a (O H)_{2}$ is $x \times 10^{- 1} N$ , where $x$ is: