Question

100 mL of sulphuric acid solution (sp. gr. =1.84) contains 98% by weight of pure acid. Calculate the volume of 0.46 M NaOH solution (in L) required to just neutralize the above acid solution.

Answer 1

Density= mass/volume

$\therefore$ mass=$density\times volume$

given density of $98$% $H_{2}S{O}_4=1.84g/mL$

$\therefore$ $1mL$ of $98$% $H_{2}S{O}_4$ will weigh

$\Rightarrow 1.84\times 1=1.84g$

$\therefore 100mL$ will weigh ##1.84\times 100=184g$

$\therefore$ concentration is given as $98$% by mass, $184g$ of the solution will contain $0.98\times 184=180.32g$ of $H_{2}S{O}_4$

molar mass of $H_{2}S{O}_4=98.1g$

$\therefore 180.32g$ constitutes $1.84$ moles

The reaction between $NaOH$ and $H_{2}S{O}_4$ is as follows:

$2NaOH+H_{2}S{O}_4⟶N{a}_{2}S{O}_4+2H_{2}O$

$\therefore$ , to neutralize $1$ mole of $H_{2}S{O}_4$,

$2$ moles of $NaOH$ are needed,

So, to neutralize $1.84$ moles,

$1.84\times 2=3.68$ moles of $NaOH$

molar mass of $NaOH=40g$

So, $3.68$ moles constitute $147g$ of $NaOH$

We know that $1L$ of $1M$ $NaOH$ contains $40g$ of $NaOH$

So, $1L$ of $0.46M$ $NaOH$ contains $18.4g$ of $NaOH$

$\therefore$ to get $147g$ of $NaOH$, we need to take $147/18.4=7.826L$ of $0.1M$ $NaOH$

$\therefore$ Volume of $0.46M$ $NaOH$ solution required to neutralize the above acid solution is $7.826L$