Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0136 M solution. The pH of the resulting solution is 2.45. Calculate the Ka for the acid.

Answer 1

Answer:
`K_a=1.2* 10^(-6)`

Explanation:
`HA\rightarrow H^++A^-`

initially conc. c 0 0

At eqm.
`c(1-\alpha)`
`c\alpha`
`c\alpha`

The expression for dissociation constant is,

`K_a=(c\alpha* c\alpha)/(c(1-\alpha))`

And,

`[H^+]=c\alpha`

`pH=-log[H^+]=2.45`

`H^+=3.5* 10^(-3)`

`3.5* 10^(-3)=0.0136\alpha`

`\alpha=0.26`

Now put all the given values in this expression, we get

`K_a=((0.0136* 0.26)^2)/(0.0136(1-0.26))`

`K_a=1.2* 10^(-6)`