JEE MAIN - Chemistry (2007 - No. 5)
The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of
the acid is ionized is :
7.0
4.5
2.5
9.5
Explanation
For acidic buffer $$pH = p{K_a} + \log \left[ {{{salt} \over {acid}}} \right]$$
or $$\,\,\,pH = p{K_a} + \log {{\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}$$
Given $$p{K_a} = 4.5\,\,$$ and acid is $$50\% $$ ionised.
$$\left[ {HA} \right] = \left[ {{A^ - }} \right]\,\,\,$$ (when acid is $$50\% $$ ionised)
$$\therefore$$ $$\,\,\,pH = p{K_a} + \log \,1$$
$$\therefore$$ $$\,\,\,pH = p{K_a} = 4.5$$
$$pOH = 14 - pH$$
$$ = 14 - 4.5$$
$$ = 9.5$$
or $$\,\,\,pH = p{K_a} + \log {{\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}$$
Given $$p{K_a} = 4.5\,\,$$ and acid is $$50\% $$ ionised.
$$\left[ {HA} \right] = \left[ {{A^ - }} \right]\,\,\,$$ (when acid is $$50\% $$ ionised)
$$\therefore$$ $$\,\,\,pH = p{K_a} + \log \,1$$
$$\therefore$$ $$\,\,\,pH = p{K_a} = 4.5$$
$$pOH = 14 - pH$$
$$ = 14 - 4.5$$
$$ = 9.5$$
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