Question

Formic acid (HCOOH), Ka = 1.8 X 10-4, is the irritant associated with ant stings. Calculate the equilibrium concentrations of all species, HA, H+, and A-1, of an aqueous solution containing 18.0 g of formic acid per liter. ?

Answer 1

We interrogate the reaction...

`HCO_2H(aq) + H_2O(l)rightleftharpoons HCO_2^(-) + H_3O^+`

Explanation:

For which....

`K_a=([HCO_2^(-)][H_3O^+])/([HCO_2H])=1.80xx10^-4`

Now to start...`[HCO_2H(aq)]=((18.0*g)/(46.03*g*mol^-1))/(1.00*L)=0.391*mol*L^-1`

And we assume that `x*mol*L^-1` of the acid undergoes protonolysis....so that...

`K_a=x^2/(0.391*mol*L^-1-x)=1.80xx10^-4`...

..and we make the usual approx. that `0.391*mol*L^-1">>"x`..

And so `x_1=sqrt(K_axx0.391)=sqrt(1.80xx10^-4xx0.391)=8.38xx10^-3*mol`

....and as usual, now we gots an approx. for `x`, we recycle this back in..

`x_2=sqrt(1.80xx10^-4xx(0.391-8.38xx10^-3))=8.30xx10^-3`

`x_3=8.30xx10^-3*mol*L^-1`...and since the approximations have converged, I am prepared to accept this as the true value.

But `x=[H_3O^+]=[H_3CO_2^(-)]=8.30xx10^-3*mol*L^-1`

And ...........

`[H_3CO_2H]=(0.391-8.30xx10^-3)*mol*L^-1=0.383*mol*L^-1`

`pH=-log_10(8.30xx10^-3)=-(-2.08)=2.08.`