The conductivity ofa solution may be taken to be directly proportional to the total concentration of the charge carriers (ions) present in it in many cases. Using the above, find the percent decrease in conductivity (k) of a solution of a week monoacidic base BOH when its 0.1 M solution is diluted to double its original volume. `(K_(b) = 10^(-5) for BOH) (take `sqrt(50) = 7.07`) (mark the answer to nearest integer)

Answer

29

Explanation

Initially `[OH^(-)]=sqrt(10^(-5)xx0.1)=10^(-3)`
`[ions]_("total")=2xx10^(-3) M`
later `[OH^(-)]=sqrt(10^(-5)xx1/20)=sqrt50xx10^(-4) M`
`[ions]_("total")=2sqrt50xx10^(-4) M`
`therefore` % change on `["ions"]_("total")=(2sqrt50-20)/20xx100=-29.29%`