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 3 of 5: Bicarbonate pH and the new buffer

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\text{(i) } \mathrm{pH} = \tfrac{1}{2}(pK_{a1} + pK_{a2}) = \tfrac{1}{2}(6.35 + 10.33) = 8.34;\qquad \text{(ii) } \mathrm{pH} = pK_{a1} + \log\frac{[\ce{HCO3^-}]}{[\ce{H2CO3}]} = 6.35 + \log\tfrac{0.021}{0.003} = 7.20

Analysis

(i) $\ce{HCO3^-}$ is amphoteric, so its pH is the mean of the two p $K_a$ values. (ii) Adding HL consumes $3.00\times10^{-3}$ mol $\ce{HCO3^-}$ and makes the same amount of $\ce{H2CO3}$ : the buffer ratio becomes 0.021/0.003, giving pH = 7.20.