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 (

[\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 (

pK_{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 (

pK_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;

pK_{a1} = 1.70

pK_{a2} = 2.60

pK_{a3} = 6.30

pK_{a4} = 10.60

)? Answer Yes or No; if yes, calculate the concentration.

Step 4 of 4: EDTA solution condition

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c = [\ce{H4A}] + [\ce{H3A-}] + [\ce{H2A^{2-}}] = [\ce{H+}],\ [\ce{H+}] = [\ce{H3A-}] + 2[\ce{H2A^{2-}}] \Rightarrow [\ce{H4A}] = [\ce{H2A^{2-}}];\quad \mathrm{pH} = \frac{pK_1+pK_2}{2} = 2.15,\ c = 0.0071\ \text{M}

Analysis

In acidic medium the 3rd and 4th steps can be neglected. Equating $c = [\ce{H4A}] + [\ce{H3A-}] + [\ce{H2A^{2-}}] = [\ce{H+}]$ with $[\ce{H+}] = [\ce{H3A-}] + 2[\ce{H2A^{2-}}]$ forces $[\ce{H4A}] = [\ce{H2A^{2-}}]$ . Then $[\ce{H+}]^2 = K_{a1}K_{a2}$ , giving pH = $(1.70+2.60)/2 = 2.15$ ( $[\ce{H+}] = 7.08\times10^{-3}$ M). Hence $[\ce{H+}] = c = 0.0071$ mol dm $^{-3}$ . Answer: Yes.