Calculate pH of a 1.0 mol L -1 boric acid solution.

Question

(Ka(H3BO3) = 5.8 x 10 power -10

Please solve this problem. My computer does not have the power symbol.

Can you please show maths and steps.
Is mollar mass required of the above.
Is -log and constant above used.

Thanks

Answer 1

H3BO3 ==> H^+ + H2BO3^-

Set up an ICE chart, substitute into Ka expression, and solve for H^+. Then use pH = -log(H^+) to convert to pH.
At equilibrium, you should have this substituted into Ka.
5.8 x 10^-10=(H^+)(H2BO3^-)/(H3BO3)
5.8 x 10^-10 = (x)(x)/(1-x)
Solve for x, which is H^+, then convert to pH.