Chemistry Labs

Problem 1

A damaged label on an acid bottle. The label on a bottle containing a dilute aqueous solution of an acid became damaged so that only its concentration was readable. A pH meter showed that the hydrogen-ion concentration was equal to the value on the label ([HX+]=c[\ce{H+}] = c). (1.1) Give the formulae of four acids that could have been in the solution if the pH changed by one unit after a tenfold dilution. (1.2) Could it be possible that the dilute solution contained sulfuric acid (pKa2=1.99pK_{a2} = 1.99)? Answer Yes or No; if yes, calculate or estimate the pH. (1.3) Could it be possible that the solution contained acetic acid (pKa=4.76pK_a = 4.76)? Answer Yes or No; if yes, calculate or estimate the pH. (1.4) Could it be possible that the solution contained EDTA (ethylenediaminetetraacetic acid; pKa1=1.70pK_{a1} = 1.70, pKa2=2.60pK_{a2} = 2.60, pKa3=6.30pK_{a3} = 6.30, pKa4=10.60pK_{a4} = 10.60)? Answer Yes or No; if yes, calculate the concentration.
Step 3 of 4: Acetic acid at ultra-dilution
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c=[HA]+[AX−]=[HX+], [HX+]=[AX−]+[OHX−]⇒[HA]=[OHX−];[HX+]3=KaKw+Ka[HX+]2⇒[HX+]=5.64×10−7 M (pH=6.25)c = [\ce{HA}] + [\ce{A-}] = [\ce{H+}],\ [\ce{H+}] = [\ce{A-}] + [\ce{OH-}] \Rightarrow [\ce{HA}] = [\ce{OH-}];\quad [\ce{H+}]^3 = K_a K_w + K_a[\ce{H+}]^2 \Rightarrow [\ce{H+}] = 5.64\times10^{-7}\ \text{M}\ (\mathrm{pH} = 6.25)
Analysis

Equating mass balance c=[HA]+[AX−]=[HX+]c = [\ce{HA}] + [\ce{A-}] = [\ce{H+}] and charge balance [HX+]=[AX−]+[OHX−][\ce{H+}] = [\ce{A-}] + [\ce{OH-}] forces [HA]=[OHX−][\ce{HA}] = [\ce{OH-}]. Substituting into Ka=[HX+][AX−]/[HA]K_a = [\ce{H+}][\ce{A-}]/[\ce{HA}] gives Ka=[HX+]([HX+]−[OHX−])/[OHX−]K_a = [\ce{H+}]([\ce{H+}] - [\ce{OH-}])/[\ce{OH-}], leading to [HX+]3=KaKw+Ka[HX+]2[\ce{H+}]^3 = K_a K_w + K_a[\ce{H+}]^2. Two iterations give [HX+]=c=5.64×10−7[\ce{H+}] = c = 5.64\times10^{-7} M (pH = 6.25). Answer: Yes.