Question

Find the pH of mixing 0.1M 200ml HCOOH & 0.1M 50ml Ca(OH)2 .
Ka = 10^-4

Answer 1

Answer:

1.4

Explanation:

its a acid base reaction so

N1V1 - N2V2 = N3V3

0.1×200 - 0.1×2×50 = N3× 250

N3 = 0.04

By taking log we get ph = 1.4

Answer 2

Answer:

pH =pka =4

Explanation:

2 HCOOH + Ca(OH)2 ➡ Ca(HCOO)2 +2H2O

N1V1 N2V2 0 0

=20 millimole 5millimole 0 0

5 millimole of ca(oh)2 reacts with 10 millimole of HCOOH as per limiting reagent.

so now 10bmillimole of HCOOH is left.

and 10 millimole of both Ca(HCOO)2 and 2 H2o is formed .

so., by pH=pka+log(salt concn) / acid concn

pH= pka +log 10/10

pH = pka

and given Ka = 10^-4 so pKa= 4

therefore pH =4