Question

What will be the pH of a 0.02 M Ca(HCO)3 ?

Answer 1

Unknown on the given data...

Explanation:

You gots a solution for which `[HCO_3^(-)]=0.04*mol*L^-1`...the parent salt is `Ca(HCO_3)_2`... We need `pK_a` data for bicarbonate ion....and the webs tell me `pK_a=10.32`

And so we got...

`HCO_3^(-) + H_2O(l) rightleftharpoonsCO_3^(2-) + H_3O^+`...

....where `K_a=10^(-10.32)=([CO_3^(2-)][H_3O^+])/([HCO_3^(-)])`

...and we make the usual approximations, rounding up the usual suspects...and `x*mol*L^-1` bicarbonate dissociates..

`K_a=10^(-10.32)=(x^2)/(0.02-x)`..

...and so if `0.02">>"x`

`x_1~=sqrt{10^(-10.32)xx0.02}=9.78xx10^-7`

`x_2~=sqrt{10^(-10.32)xx0.02}=9.78xx10^-7`

And so `[H_3O^+]=9.78xx10^-7*mol*L^-1`

`pH=-log_10(9.78xx10^-7)=6.0`