Question

The heat produced by complete neutralisation of $100ml$ of $HN{O}_3$ with $300ml$ of decimolar $KOH$ solution is $1.713kJ$. The molarity of $HN{O}_3$ solution will be:

• A. $0.1M$
• B. $1M$
• C. $0.3M$
• D. $0.5M$

Answer 1

The correct option is C $0.3M$

Given,

Complete neutralization of $100mL$ $HN{O}_3$

with $300mL$ $\frac{1}{10}M$ $KOH$ solution

produced heat=$1.713KJ$

Strong reagents generally go full ionization

$HN{O}_3+KOH⟶KN{O}_3+H_{2}O$

We are provided $300mL$ $\frac{1}{10}M$ $KOH$

$\therefore$ No. of moles of $O{H}^{-}$ produced

$=M_{KOH}\times V_{KOH}$

$=\frac{1}{10}\times 300\times {10}^{-3}$

$=0.03$ moles

Here No. of moles of $O{H}^{-}$ produced

=No. of moles of $H^{+}$

$\therefore$ Moles of $HN{O}_3=0.03$

$\Rightarrow$ Molarity of $HN{O}_3=0.03\times \frac{1000}{100}$

$=0.3M$