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Which of the following solution will have $P^{H}$ close to one?

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$100 m l$ of $\frac{M}{100} H C l + 100 m l$ of $\frac{M}{10} N a O H$

$55 m l$ of $\frac{M}{10} H C l + 45 m l$ of $\frac{M}{10} N a O H$

$10 m l$ of $\frac{M}{10} H C l + 90 m l$ of $\frac{M}{10} N a O H$

$75 m l$ of $\frac{M}{5} H C l + 25 m l$ of $\frac{M}{5} N a O H$

answer is D.

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Detailed Solution

a) is exact neutralization hence $P^{H} = 7$

b) $[H^{+}] = 10^{- 2}$ $P^{H} = 2$
c) $P^{H} > 7$
d) $[H^{+}] = 10^{- 1} P^{H} = 1$

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