.......KOH + HA ==> KA + H2O
I.....0.23..0.45....0....0
C....-0.23.-0.23...0.23..0.23
E......0....0.22...0.23
(KA) = 0.23 mols/0.680 L = ?M
(HA) = 0.22 mols/0.680 L = ?M
pH = pKa + log (base)/(acid)
pH = ?
A buffer solution is made by dissolving 0.45 moles of a weak acid (HA) and 0.23 moles of KOH into 680 mL of solution. What is the pH of this buffer? Ka = 2.4×10−6 for HA.
2 answers
5.63