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