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 3 of 3: Equal absorbances
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εHA/εAX−=1.00×10−2−3.81×10−31.69×10−3−1.00×10−3=9.0\varepsilon_{\ce{HA}}/\varepsilon_{\ce{A-}} = \dfrac{1.00\times 10^{-2} - 3.81\times 10^{-3}}{1.69\times 10^{-3} - 1.00\times 10^{-3}} = 9.0
Analysis

By Beer–Lambert, A1=l(εAX−[AX−]+εHA[HA])A_1 = l(\varepsilon_{\ce{A-}}[\ce{A-}] + \varepsilon_{\ce{HA}}[\ce{HA}]) before and after. Equating: 10−2εAX−+10−3εHA=3.81×10−3εAX−+1.69×10−3εHA10^{-2}\varepsilon_{\ce{A-}} + 10^{-3}\varepsilon_{\ce{HA}} = 3.81\times 10^{-3}\varepsilon_{\ce{A-}} + 1.69\times 10^{-3}\varepsilon_{\ce{HA}}, which rearranges to εHA/εAX−=6.19×10−3/0.69×10−3=9.0\varepsilon_{\ce{HA}}/\varepsilon_{\ce{A-}} = 6.19\times 10^{-3}/0.69\times 10^{-3} = 9.0.