Question

What is the ph of 0.51 m ca(ch3coo)2 solution?

Answer 1

Answer : The pH of calcium acetate solution will be 9.38.

Explanation : When calcium acetate dissocates completely in solution to form Calcium(+2) ions and two acetate ions, this gives the acetate ion concentration as (2 X 0.51 M) = 1.02 M

Acetate ion then hydrolyzes as follows;

`C_(2)H_(3)O_(2)^(-) + HOH <----> HC_(2)H_(2)O_(2) + OH^(-)`:

Now, Kb will be =
`[HC_(2)H_(2)O_(2)] [OH^(-)]` /
`[C_(2)H_(3)O_(2)^(-)]`

= Kw / Ka

= 1 x
`10^(-14)` / 1.8 x
`10^(-5)` = 5.6 ×
`10^(-10)`

`[OH^(-)]` = 2.4 ×
`10^(-5)`

Now, we know , pOH = -log
`[OH^(-)]` = 4.62

So, the pH = 14 - pOH = 14 - 4.62 = 9.38