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प्रश्न
Answer the following in brief :
The pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation constant.
उत्तर
Given:
pH = 3.2
Concentration Dissociation = 0.02 M
Dissociation constant (Ka) = ?
Calculation:
Ka = `(Calpha^2)/(1 - alpha)`
`alpha = {[H^+]}/C`
pH = −log10[H+]
3.2 = −log10[H+]
−3.2 = log10[H+]
`10^-3.2 = [H^+]`
`6.3 xx 10^-4 = [H^+]`
`alpha = {[H^+]}/C`
= `(6.3 xx 10^-4)/0.02`
= 0.0315
Ka = `(Calpha^2)/(1 - alpha)`
`(0.02 xx (0.0315)^2)/(1 - 0.0315)`
= `(0.02 xx 0.000992)/0.9685`
= `(0.02 xx 9.92 xx 10^-4)/(1)`
= `0.1984 xx 10^-4`
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