Chemistry Labs

Problem 3

Ignore the absorption of the cell and the solvent; all solutions are at 25 °C. An aqueous solution X was prepared from the weak acid HA and NaA; in X the concentrations are [AX−]=1.00×10−2[\ce{A-}] = 1.00\times 10^{-2}, [HA]=1.00×10−3[\ce{HA}] = 1.00\times 10^{-3} and [HX+]=1.00×10−4[\ce{H+}] = 1.00\times 10^{-4} mol L−1^{-1}, linked by the equilibrium HA⇌HX++AX−\ce{HA <=> H+ + A-}. The absorbance of X was A1A_1 at wavelength λ1\lambda_1 (optical path ll). Then X was diluted to twice its initial volume using hydrochloric acid of pH 2.500; after the dilution and re-establishment of equilibrium, the absorbance at λ1\lambda_1 was still A1A_1. Determine the ratio εHA/εAX−\varepsilon_{\ce{HA}}/\varepsilon_{\ce{A-}} of the absorption coefficients of HA and AX−\ce{A-} at λ1\lambda_1.
Step 2 of 3: New equilibrium after dilution
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[HX+]′=10−2.500=3.16×10−3;[AX−]′=5.00×10−3−x=3.81×10−3, [HA]′=5.00×10−4+x=1.69×10−3 mol L−1[\ce{H+}]' = 10^{-2.500} = 3.16\times 10^{-3};\quad [\ce{A-}]' = 5.00\times 10^{-3} - x = 3.81\times 10^{-3},\ [\ce{HA}]' = 5.00\times 10^{-4} + x = 1.69\times 10^{-3}\ \text{mol L}^{-1}
Analysis

Dilution halves the analytical concentrations ([HA]=5.00×10−4[\ce{HA}] = 5.00\times 10^{-4}, [AX−]=5.00×10−3[\ce{A-}] = 5.00\times 10^{-3}) while HCl fixes [HX+]′=3.16×10−3[\ce{H+}]' = 3.16\times 10^{-3} mol L−1^{-1}. Enforcing Ka=[HX+]′[AX−]′/[HA]′=10−3K_a = [\ce{H+}]'[\ce{A-}]'/[\ce{HA}]' = 10^{-3} gives x=1.19×10−3x = 1.19\times 10^{-3}, i.e. [AX−]′=3.81×10−3[\ce{A-}]' = 3.81\times 10^{-3} and [HA]′=1.69×10−3[\ce{HA}]' = 1.69\times 10^{-3} mol L−1^{-1}.

Common pitfall. The diluting HCl supplies extra H+, so the equilibrium shifts toward HA; do not simply halve the equilibrium concentrations.