Question

Find the percentage ionization of 0.001 M leucine hydrochloride if pKa1= 2.40 pKa2 = 9.60 please show me how to find it?

Answer 1

Answer on Question#69722 – Chemistry – Organic chemistry

Question: Find the percentage ionization of 0.001 M leucine hydrochloride if pKa1 = 2.40 pKa2 = 9.60 please show me how to find it?

Solution:

$$\mathrm{H_2L^+ = HL + H^+}$$ $$\mathrm{Ka_1 = 10^{-pKa1} = 3.98 \times 10^{-3}}$$ $$\mathrm{Ka_2 = 10^{-pKa2} = 2.5 \times 10^{-10}}$$

$\mathrm{Ka_1 >> Ka_2}$ and the percent ionization is determined just the first stage of ionization.

$$0.00398 = \frac{x^2}{0.001 - x}$$ $$x^2 = 0.00398(0.001 - x)$$ $$x^2 + 0.00398x - 0.00000398 = 0$$ $$x = 8.28 \times 10^{-4}$$ $$\text{percent ionization} = \frac{8.28 \times 10^{-4}}{0.001} \times 100\% = 82.8\%$$

Answer: 82.8%

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