H3BO3 is a weak acid because it ionizes to provide H^+ in solution. Ka1 = 7.2E-10. Ka2 is 1.8E-13 but for pH of this solution need not be considered since it is 1000 times weaker than Ka1 and contributes very little on its own.
................H3BO3 ==> H^+ + [H2BO3]^-
I..............0.325............0..............0
C................-x...............x...............x
E.............0.325-x..........x...............x
Plug the E line into the Ka expression and solve for x = (H^+), then convert this to pH by pH = -log*H^+).
Post your work if you get stuck.
What is the pH of a 100.0 mL 0.325 M H3BO3 solution at 25 C? (Why is this an acidic solution?)
1 answer