Answer:
The 91.8% is present as bicarbonate ion
Explanation:
It is possible to find pH of a buffer using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is 7.40; pKa is 6.36, [A⁻] is concentration of bicarbonate ion and [HA] concentration of carbonic acid</em>
7.40 = 6.35 + log [A⁻] / [HA]
1.05 = log [A⁻] / [HA]
11.22 = [A⁻] / [HA] <em>(1)</em>
As the question is in therms of percentage:
[A⁻] + [HA] = 100 <em>(2)</em>
Replacing (2) in (1):
11.22 = [A⁻] / 100 - [A⁻]
1122 - 11.22[A⁻] = [A⁻]
1122 = 12.22[A⁻]
91,8% = [A⁻]
<h3>The 91.8% is present as bicarbonate ion</h3>
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