Question

What is the pH of a 0.25 M aqueous solution of KCHO2 at 25°C?

The Ka of HCHO2 is1.8 x 10-4 at 25°C.

Answer 1

kCHO2 +H2O↔HCHO2+H3O+

ka=[HCHO2][H3O+]/[KCHO2]

ICE table

	[KCHO2]	[HCHO2]	[H3O+]
Initial	0.25M	0	0
change	-x	+x	+x
equibrium	0.25-x	x	x

Ka=1.8*10^-4=x*x/0.25-x (ignore x <<<0.25 as the value of ka is very small so dissociation is less)

1.8*10^-4=x*x/0.25

X^2=0.45*10^-4

X=0.67*10^-2=0.0067 M=[HCHO2]

[KCHO2]=0.25-0.0067=0.2433M

Using henderson -hasselbach equation,

pH=pka+log[base]/[acid]=pka+log(0.0067/0.2433)

pka=-logka=-log(1.8*10^-4)=4-0.25=3.75

pH=3.75+log 0.0275

     =3.75+(-1.56)=2.19

pH=2.19