Question

Calculate the percent ionization of 0.140 m lactic acid in a solution containing 8.5×10−3 m sodium lactate.

Answer 1

Chemical reaction 1: C₃H₆O₃(aq) ⇄ C₃H₅O₃⁻(aq) + H⁺(aq).
Chemical reaction 2: C₃H₅O₃Na(aq) → C₃H₅O₃⁻(aq) + Na⁺(aq).
c(lactic acid - C₃H₆O₃) = 0,140 M.
c(sodim lactate - C₃H₅O₃Na) = 8,5·10⁻³ M.
Ka(C₃H₆O₃) = 1,4·10⁻⁴.
[H⁺] = x.
[C₃H₅O₃⁻] = 0,0085 M + x.
[C₃H₆O₃] = 0,140 M - x.
Ka = [C₃H₅O₃⁻] · [H⁺] / [C₃H₆O₃].
1,4·10⁻⁴ = (0,0085 M + x) · x / (0,140 M - x).
1,4·10⁻⁴ = 0,0085x + x² / (0,140 - x).
x = 0,00188 M.
percent ionization of lactic acid: 
α = 0,00188 M ÷ 0,140 M · 100% = 1,35%.

Answer 2

When an atom or the molecules gains and losses electrons to become positive and negatively charged are called ionization. The per cent ionization of lactic acid is 1.35%.

What is percentage ionization?

Percentage ionization is the amount of the ionized atoms or the molecules in the terms of the per cent. The method of the molecule or an atom to acquire the electron or to lose an electron to become negative or positive in charge is called ionization.

The chemical reactions can be shown as,

$\rm C_{3}H_{6}O_{3}(aq)\rightleftharpoons C_{3}H_{5}O_{3}^{-}(aq) + H^{+}(aq)$

$\rm C_{3}H_{5}O_{3}Na(aq) \rightarrow C_{3}H_{5}O_{3}^{-}(aq) + Na^{+}(aq)$

Given,

• Concentration of lactic acid = 0.140 M
• The concentration of sodium lactate = $8.5 \times 10^{-3} \;\rm M$
• Ka of lactic acid = $1.4 \times 10^{-4}$
• The concentration of hydrogen ion $\rm [H^{+}]$ = x

Acid dissociation constant can be shown as,

$\begin{aligned} \rm Ka &= \rm \dfrac{[C_{3}H_{5}O_{3}^{-}] \times [H^{+}] }{[C_{3}H_{6}O_{3}]}\\\\1.4 \times 10^{-4} &= \rm \dfrac{(0.0085 M + x) \times x }{ (0.140 M - x)}\\\\&= 0.00188\;\rm M\end{aligned}$

Percentage ionization:

$\begin{aligned}\alpha &= \dfrac{0.00188\;\rm M}{0.140 \;\rm M} \times 100\%\\\\&= 1.35\%\end{aligned}$

Therefore, per cent ionization is 1.35 %.

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