...........HF + NaOH ==> NaF + H2O
begin....0.6 mol..0.12...1.06
change..-0.12......-0.12.+0.12
equil...0.48........0.....1.18
Use the Henderson-Hasselbalch equation
pH = pKa + log [(base)/(acid)]
NaF is the base. HF is the acid. pKa is that of HF.
Calculate the pH after 0.12 mol of NaOH is added to 1.00 L of 0.60 M HF and 1.06 M KF
1 answer