Calculate the equilibrium concentrations of boric acid H3BO3

Calculate the equilibrium concentrations of boric acid (H3BO3) and the borate anion (H2BO3-) in pH 8 seawater, given a total boron concentration (H3BO3 + H2BO3-) = 4.2 * 10-4 mol L -1 and pK = 8.6. Assume ideal conditions, that the activity coefficients are one.

Solution

let the concentration of the base be x.so, acid=4.2*10^-4-x

pH=pKa log([Base]/[Acid])
8=8.6 log(x/(4.2*10^-4-x))...solving we get x=8.34*10^-5 M

H3BO3--->H+ + H2BO3 -

SO,

10^-8.6= 10^-8*x/(4.2*10^-4 -x)

or x=8.34*10^-5 M

so, [H2BO3 -]=8.34*10^-5 M

[H3BO3]=3.366*10^-4 M