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The `pK_(a)` 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:A. `4.5`B. `7.0`C. `9.5`D. `2.5` |
Answer» Correct Answer - C Ionization of weak acid (HA) is represented as `HA(aq.)hArr H^(+)(aq.)+A^(-)(aq.)` When HA is `50%` ionized. `C_(HA)=C_(A^(-))` According to the Henderson equation, `pH=pK_(a)+log.(C_(A^(-)))/(C_(HA))` `=pK_(a)+log 1` `=pK_(a)=4.5` At `25^(@)C`, `pH+pOH=pK_(w)=14` Thus, `pOH=14-pH` `=14-4.5` `=9.5` |
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