Question

Citric acid is found in lemon juice.
$H{O}_{2}CC{H}_{2}C(OH)(C{O}_{2}H)C{H}_{2}C{O}_{2}H$ (citric acid)
What is the volume of 0.4 mol $d{m}^{-3}$ sodium hydroxide solution required to neutralise a solution containing 0.005 mol of citric acid ?
a.) 12.5$c{m}^3$
b.) 25.0$c{m}^3$
c.) 37.5$c{m}^3$
d.) 50.0$c{m}^3$

Answer 1

Hint: The sodium hydroxide is a strong base. It reacts with acids to produce salt and water, thereby, neutralising the acid. Only one mole of NaOH can neutralise one $H^{+}$ion because it has one hydroxide ion to neutralise. One mole of citric acid can release three $H^{+}$ions. So, it requires three moles of sodium hydroxide for one mole of citric acid.

Complete answer:
Let us write what is given to us and what we need to find out.
Given :
Concentration of sodium hydroxide solution = 0.4 mol$d{m}^{-3}$
Molarity of Citric acid = 0.005 mol
To find :
Volume of NaOH solution required
We know that normally 1 mole of NaOH neutralize 1 mole of $H^{+}$ion.
Further, we have structure of citric acid as -
$H{O}_{2}CC{H}_{2}C(OH)(C{O}_{2}H)C{H}_{2}C{O}_{2}H$
There are three $H^{+}$ions that are released by 1 mole of citric acid. So, this means 3 moles of sodium hydroxide will be required to neutralise 1 mole of citric acid.
We have 0.005 moles of citric acid to neutralise.
Thus, moles of NaOH required = 3$\times$0.005
Moles of NaOH required = 0.015 mol
We have the formula for number of moles as -
Number of moles = Concentration$\times$Volume
Thus, volume of NaOH required = ${\displaystyle \frac{Number\text{ of moles}}{Concentration}}$
So, volume of NaOH required = ${\displaystyle \frac{0.015}{0.4}}$
volume of NaOH required = 0.375$d{m}^3$
volume of NaOH required = 37.5$c{m}^3$

So, the correct option is option c.).

Note:
It must be noted that the number of moles can be calculated by product of concentration with volume of sodium hydroxide solution. The sodium hydroxide has one hydroxide ion while the citric acid has three protons.