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

Hint: 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.