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 4 of 5: Lactic acid added to blood

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\frac{[\ce{HCO3^-}]_B}{[\ce{H2CO3}]_B} = \frac{K_{a1}}{10^{-7.00}} = 4.5,\ \ [\ce{HCO3^-}]_B + [\ce{H2CO3}]_B = 0.0239\ \Rightarrow\ \Delta[\ce{H2CO3}] = 2.4\times10^{-3}\ \mathrm{mol}

Analysis

At pH 7.40 the buffer ratio is $10^{-6.35}/10^{-7.40} = 11.3$ , giving $[\ce{H2CO3}]_A = 0.0019$ M and total carbonate 0.0239 M. At pH 7.00 the ratio is 4.5; solving the two equations gives $[\ce{H2CO3}]_B = 0.0043$ M. Each mole of HL creates one mole of $\ce{H2CO3}$ , so $n(HL) = 2.4\times10^{-3}$ mol per litre.