1.

A monobasic weak acid solution has a molarity of 0.005 and pH of 5. What is the percentage ionization in this solution ?

Answer»

`2.0`
`0.2`
`0.5`
`0.25`

SOLUTION :`HA to H^(+) + A^(-)`
`K_(a)=([H^(+)][A^(-)])/([HA]), because [H^(+)]=10^(-pH)`
`therefore [H^(+)]=10^(-5)`, and at equilibrium `[H^(+)]=[A^(-)]`
`therefore K_(a)=(10^(-5)xx10^(-5))/(0.0015)=2xx10^(-8)`
`alpha= sqrt((K_(a))/(005))=sqrt(4xx10^(-6))=2xx10^(-3)`
PERCENTAGE IONIZATION = 0.2


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