Ionization constant of 0.1M bromo acetic acid is 0.132. Then calculate pH and pKa of this solution
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Ionization constant of 0.1M bromo acetic acid is 0.132. Then calculate pH and pKa of this solution
Ionization constant of 0.1M bromo acetic acid is 0.132. Then calculate pH and pKa of this solution
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[tex] \huge \rm {Answer:-} [/tex]
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[tex] \large \sf \blue {Given:-} [/tex]
[tex] \to \tt {Concentration\: (C)=0.1M} [/tex]
[tex] \to \tt {Degree\: of\: ionisation\: (\alpha)=0.132} [/tex]
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[tex] \large \sf \orange {To\: Calculate:-} [/tex]
[tex] \to \tt {PH\: of\: Bromo\: acetic\: acid=?} [/tex]
[tex] \to \tt{PK_{a}\: of\: Bromo\: acetic\: acid=?} [/tex]
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[tex] \large \sf \purple {We\: Know,} [/tex]
[tex]\large \implies \bf {PH=-log[H^{+}]} [/tex]
[tex]\implies \bf {[H^{+}]=?} [/tex]
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[tex]\small \implies \bf {[H^{+}]=Concentration \times Degree\: of\: ionisation} [/tex]
[tex]\implies \bf {[H^{+}]=C\times\alpha} [/tex]
[tex]\implies \bf {[H^{+}]=0.1M \times 0.132} [/tex]
[tex]\implies \bf {[H^{+}]=0.0132M} [/tex]
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[tex] \large \sf \pink {Now,} [/tex]
[tex]\large \to \bf {PH=-log[H^{+}]} [/tex]
[tex]\large \to \bf {PH=-log(0.0132)} [/tex]
[tex]\large \to \bf {PH=1.879} [/tex]
[tex] \bf \to {Which\: is\: nearly\: equal\:to,} [/tex]
[tex] \large \implies \bf {\fbox{PH=1.88}} [/tex]
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[tex]\large \to \bf {PK_{a} } [/tex]
[tex]\large \to \bf {PK_{a} =-logk_{a}} [/tex]
[tex]\small \to \bf {K_{a}=Concentration\times Degree\: of\: ionisation^{2} } [/tex]
[tex] \to \bf {K_{a}=C\times \alpha^{2} } [/tex]
[tex] \to \bf {K_{a}=0.1M\times (0.132)^{2} } [/tex]
[tex] \to \bf {K_{a}=0.1M\times 0.0174} [/tex]
[tex] \to \bf {K_{a}=0.00174} [/tex]
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[tex] \large \sf \red {Accordingly,} [/tex]
[tex]\large \to \bf {PK_{a} =-logk_{a}} [/tex]
[tex]\large \to \bf {PK_{a} =-log(0.00174)} [/tex]
[tex]\large \to \bf {PK_{a} =2.769} [/tex]
[tex]\small \bf \to {So,\: Considering\: aprrox\:value,} [/tex]
[tex] \large \implies \sf {\fbox{2.7}} [/tex]
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[tex] \large \sf \green {Thenceforth,} [/tex]
[tex]\small\to \tt {PH\: of\: Bromo\: acetic\: acid=1.88} [/tex]
[tex]\small \to \tt {PK_{a}\: of\: Bromo\: acetic\: acid=2.7} [/tex]
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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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