Problem 1

Lactic acid (HL, monoprotic,

K_{a}(HL) = 1.4\times10^{-4}

) is formed in muscles during intense activity; in blood it is neutralised by hydrogen carbonate. Carbonic acid has

K_{a1} = 4.5\times10^{-7}

and

K_{a2} = 4.7\times10^{-11}

; all

\ce{CO2}

stays dissolved. (1.1) Calculate the pH of a

3.00\times10^{-3}

M solution of HL. (1.2) Calculate the equilibrium constant of

\ce{HL + HCO3^- -> H2CO3 + L^-}

. (1.3)

3.00\times10^{-3}

mol of HL is added to

1.00\ \mathrm{dm^3}

of 0.024 M

\ce{NaHCO3}

(no volume change, HL fully neutralised): (i) pH of the

\ce{NaHCO3}

solution before addition? (ii) pH after addition? (1.4) Blood at pH 7.40 with

[\ce{HCO3^-}] = 0.022

M falls to pH 7.00 after exercise: how many moles of HL were added per litre? (1.5) In saturated

\ce{CaCO3}

solution pH = 9.95; calculate the solubility and show

K_{sp} \approx 5\times10^{-9}

. (1.6) Find the maximum free

[\ce{Ca^{2+}}]

in the pH 7.40 blood of part 1.4.

Step 1 of 5: pH of the lactic acid solution

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K_a = \frac{x^2}{c_0 - x}:\quad x = \frac{-K_a + \sqrt{K_a^2 + 4K_a c_0}}{2} = 5.8\times10^{-4}\ \Rightarrow\ \mathrm{pH} = 3.24

Analysis

With $c_0 = 3.00\times10^{-3}$ M the weak-acid approximation fails ( $x/c_0 \approx 0.2$ ), so the quadratic $x^2 + K_a x - K_a c_0 = 0$ is solved: $x = [\ce{H3O+}] = 5.8\times10^{-4}$ M, pH = 3.24.

Common pitfall. The approximation

[\ce{H+}] \approx \sqrt{K_a c_0}

gives

6.5\times10^{-4}

M — about 12 % off; always check that

x \ll c_0