Question

Consider the following balanced chemical equation:
$NaOH\left(aq\right)+HN{O}_3\left(aq\right)\to NaN{O}_3\left(aq\right)+H_{2}O\left(l\right)$
If $43.55$ grams of $NaOH$ is present in $1.23L$ of an $NaOH$ solution of unknown concentration and mixed with $1.35L$ of a $1.25$ molar $HN{O}_3$ solution, which of the following is true?

• A. The resulting solution has a pH of $7$.
• B. The resulting solution has a pH less than $7$.
• C. The resulting solution has a pH greater than $7$.
• D. The resulting solution is no longer aqueous because a species other than water is present in greatest quantity and is now the solvent.

Answer 1

Answer: (B)

The resulting solution has a pH less than $7$.

Moles of $NaOH=\frac{43.55}{40}=1.088$

Moles of $HN{O}_3=1.35\times 1.25=1.687$

Moles of $HN{O}_3>$ Moles of $NaOH$

$\therefore$ Moles of $HN{O}_3$ left after neutralization $=0.5999$

Total volume $=1.25+1.35=2.58$ $L$

$\therefore \left[H^{+}\right]=\frac{0.5999}{2.58}=0.232$

$pH=-\log \left[H^{+}\right]=0.634$

$\therefore$ The resulting solution has $pH$ less than $7$.