Question

If the heat of neutralization for a strong acid - base reaction is $- 57.1kJ$, what would be the heat released when $350c{m^3}$ at $0.20M$ ${H_2}S{O_4}$ is mixed with $650c{m^3}$ of $0.01M$ $NaOH$?
A.$37.1KJ$
B.$3.71KJ$
C.$3.17KJ$
D.$0.31KJ$

Answer 1

We have to remember that when an acid and a base react together they form a neutral salt, this type of reaction also known as neutralization reaction sometimes. In the question given above hydrosulphuric acid and sodium hydroxide being a base reacts to form a neutral salt.

Complete answer:
We need to solve the question:
${H^ + } + O{H^ - } \to {H_2}O$, Heat of neutralization is given i.e. $\Delta H = - 57.1KJ$ this indicates that one mole of ${H^ + }$ion reacts with one mole of $O{H^ - }$ ion to produce one mole of ${H_2}O$ which has heat of neutralization to be $- 57.1KJ$ .
Millimoles of hydrosulfuric acid that is a strong dibasic acid will be:
$M \times V = 0.20M \times 350$
Millimoles will be $70mmol$
Thus the amount of protons (${H^ + }$ions) in the acid will be $2 \times 70 = 140mmol$
Similarly we will calculate the millimoles of sodium hydroxide base that is a monoacidic base will be
$M \times V = 0.10M \times 650$
Millimoles will be $65mmol$
Thus the amount of hydroxide ion $\left( {O{H^ - }} \right)$in the base will be $1 \times 65 = 65mmol$
Here hydroxide ion is the limiting reactant as the value of millimoles is less than ${H^ + }$ ions, when both reacts together that means 65mmol of sulfuric acid and $65mmol$ of sodium hydroxide thus we will left with $75mmol$ of sulphuric acid in the reaction. So 65mmol of water is produced from 65mmol of sulphuric acid and 65mmol of sodium hydroxide,
Thus
$65 \times {10^{ - 3}}$ mol of $O{H^ - }$ ions will produce
$= 57.1 \times 65 \times {10^{ - 3}}$
$3.71kJ$

Note:
We have to remember that a proton and a hydroxide ion react together to produce water that is a neutralized product. The limiting reagent is always the one that has less moles; it is named so as it is limiting the reaction to produce the product.