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 5 of 5: CaCO₃ solubility and free Ca²⁺ in blood

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[\ce{OH^-}] = 10^{-4.05} = 8.9\times10^{-5};\quad [\ce{CO3^{2-}}] = \frac{[\ce{HCO3^-}][\ce{OH^-}]}{K_b} = 3.8\times10^{-5};\quad s = [\ce{Ca^{2+}}] = 1.3\times10^{-4}\ \Rightarrow\ K_{sp} \approx 5\times10^{-9};\quad [\ce{Ca^{2+}}]_{max} = \frac{K_{sp}}{K_{a2}[\ce{HCO3^-}]/[\ce{H+}]} = 1.9\times10^{-4}\ \mathrm{M}

Analysis

In saturated $\ce{CaCO3}$ , carbonate hydrolyses: $[\ce{HCO3^-}] = [\ce{OH^-}] = 8.9\times10^{-5}$ M with $K_b = K_w/K_{a2} = 2.1\times10^{-4}$ , giving $[\ce{CO3^{2-}}] = 3.8\times10^{-5}$ and $s = [\ce{Ca^{2+}}] = 1.3\times10^{-4}$ M, so $K_{sp} = 4.9\times10^{-9} \approx 5\times10^{-9}$ . In the pH 7.40 buffer, $[\ce{CO3^{2-}}] = K_{a2}[\ce{HCO3^-}]/[\ce{H+}] = 2.6\times10^{-5}$ M, capping free $\ce{Ca^{2+}}$ at $K_{sp}/[\ce{CO3^{2-}}] = 1.9\times10^{-4}$ M.