College Chemistry
posted by Elleni on .
Calculate pH of a 1.0 mol L 1 boric acid solution.
(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

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)/(1x)
Solve for x, which is H^+, then convert to pH.