Determine the concentrations of the ionic species present in a 0.268 M solution of the NaO2CCOCH2CO2Na....

Determine the concentrations of the ionic species present in a 0.268 M solution of the NaO₂CCOCH₂CO₂Na. (pK_a1=3.40, pK_a2=5.11 for HO₂CCOCH₂CO₂H).

a) [HO₂CCOCH₂CO₂H]

b)[HO₂CCOCH₂CO₂^-]

c) [^-O₂CCOCH₂CO₂^-]

d) [H₃O⁺]

e) [OH^-]

Expert Solution

Here, [NaO₂CCOCH₂CO₂Na] = 0.268 M

pH = 7 + 1/2 (pKa2 + Log[NaO₂CCOCH₂CO₂Na])

i.e. pH = 7 + 1/2 (5.11 + Log(0.268))

i.e. pH = 9.269

i.e. [H₃O⁺] = 10^-9.269 = 5.382*10^-10 M

Now, [OH^-] = 10^-14/(5.382*10^-10) = 1.858*10^-5 M

queen_honey_blossom answered 2 years ago

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