Citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\), is a triprotic acid. It occurs naturally in citrus fruits like lemons and has applications in food flavouring and preservatives. A solution containing an unknown concentration of the acid is titrated with KOH. It requires $23.20 \mathrm{~mL}\( of \)0.500 \mathrm{M} \mathrm{KOH}$ solution to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the citric acid solution.

The chemical equation for the reaction between citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\), and potassium hydroxide (KOH) can be written as: \[ 1 \times \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} + 3 \times \mathrm{KOH} \to \mathrm{K}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} + 3 \times \mathrm{H}_{2} \mathrm{O}\]

Rewrite the balanced chemical equation with the ionic species, and then remove the spectator ions to find the net ionic equation. Citric acid → \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}^{3-} + 3\mathrm{H^{+}}\) 3 KOH → 3 \(\mathrm{K^{+}} + 3\mathrm{OH^{-}}\) The net ionic equation will be: \[ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}^{3-} + 3\mathrm{H^{+}} + 3\mathrm{OH^{-}} \to \mathrm{K}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} + 3 \times \mathrm{H}_{2} \mathrm{O}\]

From the net ionic equation, we know that 3 moles of KOH neutralize 1 mole of citric acid. It's given that \(23.20 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) KOH solution is required to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. First, calculate the moles of KOH used: moles of KOH = Molarity x Volume moles of KOH = \(0.500 \mathrm{M} \times 23.20 \times 10^{-3} L\) Moles of KOH = \(0.0116 \mathrm{~mol}\) From the stoichiometry, 3 mol of KOH neutralize 1 mol of citric acid. So, we can find the moles of citric acid using this ratio: moles of citric acid = \(\frac{1}{3}\) × moles of KOH = \(\frac{1}{3} \times 0.0116 \mathrm{~mol}\) moles of citric acid = \(0.00387 \mathrm{~mol}\) Now, calculate the molarity of the citric acid solution: Molarity of citric acid solution = \(\frac{\text{moles of citric acid}}{\text{volume of solution in L}}\) Molarity of citric acid solution = \(\frac{0.00387 \mathrm{~mol}}{100.00 \times 10^{-3} L}\) Molarity of citric acid solution = \(0.0387 \mathrm{M}\) The molarity of the citric acid solution is \(0.0387 \mathrm{M}\).