8 g NaOH and 4.9 g $H_{2} {S O}_{4}$ are present in one litre of the solution . What is its pH

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Solution

The correct option is B 13
$8 g m$ of $N a O H + H_{2} {S O}_{4} \to {N a H S O}_{4} + H_{2} O$
Mole $= \frac{8}{40}$
$M = \frac{4.9}{9.8}$
So now 0.1 mole of $N a O H$ is left.
NaOH is the strong base so it dissociates completely into
ion ${O H}^{0}$
This results in 0.1 moles of $(-)^{+}$ ion $P o H = - log [0.1]$
$P H = 14 - 1$
$= 13$ $8 g m$ of $N a O H + H_{2} {S O}_{4} \to {N a H S O}_{4} + H_{2} O$
Mole $= \frac{8}{40}$
$M = \frac{4.9}{9.8}$
So now 0.1 mole of $N a O H$ is left.
NaOH is the strong base so it dissociates completely into
ion ${O H}^{0}$
This results in 0.1 moles of $(-)^{+}$ ion $P o H = - log [0.1]$
$P H = 14 - 1$
$= 13$

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