Question 7.7: Fractions of Species of a Polyprotic Acid Calculate the frac...

The fraction of each species present is the concentration of that species divided by the total concentrations of all three species. For example, for HCO _{3}^{-}   we have

Fraction HCO _{3}^{-}= \frac{[HCO _{3}^{-}]}{[H_{2}CO _{3}]  + [HCO _{3}^{-}] + [CO _{3}^{2-}]}  = f_{HCO _{3}^{-}}

Dividing the numerator and denominator by [HCO _{3}^{-}]  gives

f_{HCO _{3}^{-}} = \frac{1}{ \frac{[H_{2}CO _{3}] }{[HCO _{3}^{-}]} +1+\frac{[CO _{3}^{2-}]}{[HCO _{3}^{-}]}}

We can calculate the [H_{2}CO _{3}] / [HCO _{3}^{-}]  ratio from K_{a_1} :

K_{a_1} = \frac{[H^{+}] [HCO _{3}^{-}]}{[H_{2}CO _{3}]}

or                        \frac{[H_{2}CO _{3}]}{[HCO _{3}^{-}]} = \frac{[H^{+}]}{K_{a_1} }

Since K_{a_1} = 4.3 ×10 ^{-7}   and pH = 9.00 ( [H^{+}] = 1.00  ×10 ^{-9} ),

\frac{[H_{2}CO _{3}]}{[HCO _{3}^{-}]} = \frac{1.00  ×10 ^{-9}}{4.3 ×10 ^{-7}} = 2.3 ×10 ^{-3}

We can use K_{a_2} to calculate the [CO _{3}^{2-}] /  [HCO _{3}^{-}] ratio:

K_{a_2} = \frac{[H^{+}]  [CO _{3}^{2-}] }{[HCO _{3}^{-}]} =4.8 ×10 ^{-11}

so

\frac{ [CO _{3}^{2-}]}{[HCO _{3}^{-}]} = \frac{K_{a_2}}{[H^{+}]}= \frac{4.8 ×10 ^{-11}}{1.00 ×10 ^{-9}}= 4.8 ×10 ^{-2}

Now we can calculate the fraction of [HCO _{3}^{-}] present:

f_{HCO _{3}^{-}} = \frac{1}{ \frac{[H_{2}CO _{3}] }{[HCO _{3}^{-}]} +1+\frac{[CO _{3}^{2-}]}{[HCO _{3}^{-}]}}= \frac{1}{2.3 ×10 ^{-3} + 1+ 4.8 ×10 ^{-2} }

= \frac{1}{1.05}= 0.95

Note that at pH 9, HCO _{3}^{-} represents 95% of the carbonate-containing species in the solution.