Question

Ionization constant of 0.1M bromo acetic acid is 0.132. Then calculate pH and pKa of this solution​

Answer 1

$\huge \rm {Answer:-}$

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$\large \sf \blue {Given:-}$

$\to \tt {Concentration\: (C)=0.1M}$

$\to \tt {Degree\: of\: ionisation\: (\alpha)=0.132}$

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$\large \sf \orange {To\: Calculate:-}$

$\to \tt {PH\: of\: Bromo\: acetic\: acid=?}$

$\to \tt{PK_{a}\: of\: Bromo\: acetic\: acid=?}$

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$\large \sf \purple {We\: Know,}$

$\large \implies \bf {PH=-log[H^{+}]}$

$\implies \bf {[H^{+}]=?}$

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$\small \implies \bf {[H^{+}]=Concentration \times Degree\: of\: ionisation}$

$\implies \bf {[H^{+}]=C\times\alpha}$

$\implies \bf {[H^{+}]=0.1M \times 0.132}$

$\implies \bf {[H^{+}]=0.0132M}$

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$\large \sf \pink {Now,}$

$\large \to \bf {PH=-log[H^{+}]}$

$\large \to \bf {PH=-log(0.0132)}$

$\large \to \bf {PH=1.879}$

$\bf \to {Which\: is\: nearly\: equal\:to,}$

$\large \implies \bf {\fbox{PH=1.88}}$

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$\large \to \bf {PK_{a} }$

$\large \to \bf {PK_{a} =-logk_{a}}$

$\small \to \bf {K_{a}=Concentration\times Degree\: of\: ionisation^{2} }$

$\to \bf {K_{a}=C\times \alpha^{2} }$

$\to \bf {K_{a}=0.1M\times (0.132)^{2} }$

$\to \bf {K_{a}=0.1M\times 0.0174}$

$\to \bf {K_{a}=0.00174}$

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$\large \sf \red {Accordingly,}$

$\large \to \bf {PK_{a} =-logk_{a}}$

$\large \to \bf {PK_{a} =-log(0.00174)}$

$\large \to \bf {PK_{a} =2.769}$

$\small \bf \to {So,\: Considering\: aprrox\:value,}$

$\large \implies \sf {\fbox{2.7}}$

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$\large \sf \green {Thenceforth,}$

$\small\to \tt {PH\: of\: Bromo\: acetic\: acid=1.88}$

$\small \to \tt {PK_{a}\: of\: Bromo\: acetic\: acid=2.7}$

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NOTE:- The value of log & anti log of a number can be found out using the longarithm and anti-longarithm tables.

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