I have 5.217 which rounds to 5.22. Good work. I have
pH = 4.74 + log (12/4)
You may not have used the same units I did for base/acid but you should have taken the log of 3 no matter what units were used.
Acetic acid (HOAc) is a weak acid with a Ka= 1.8 X 10^-5 and sodium hydroxide (NaOH) is a strong base. What is the pH of a solution made by mixing 75.0 ml of a 0.160 M solution NaOH and 25.0 ml of a 0.640 M solution of HOAc?
I got a pH of 5.22, but I am not sure if this is right. Please let me know... I have a big exam tomorrow and want to make sure I am doing these correctly. Thanks.
2 answers
Net Ionic Reaction:
HC2H3O2(aq) + OH^-(aq) --> C2H3O2^-(aq) + H2O
Initial moles of oH- = (0.075L)(0.160mol/L) = 0.012 mol OH- or NaOH
Initial Moles HOAc- = (0.025L)(0.640mol/L) = 0.016 mol HOAc
Final Moles of C2H3O2^- = moles OH- (limiting reagent)= 0.012
Final moles of HOAc = 0.016 - 0.012 = 0.004
pH = pKa + log([C2H3O2^-] / [HOAc])
pKa = 4.74, [C2H3O2^-]=0.012, [HOAc]= 0.004
You decide if your answer is correct
HC2H3O2(aq) + OH^-(aq) --> C2H3O2^-(aq) + H2O
Initial moles of oH- = (0.075L)(0.160mol/L) = 0.012 mol OH- or NaOH
Initial Moles HOAc- = (0.025L)(0.640mol/L) = 0.016 mol HOAc
Final Moles of C2H3O2^- = moles OH- (limiting reagent)= 0.012
Final moles of HOAc = 0.016 - 0.012 = 0.004
pH = pKa + log([C2H3O2^-] / [HOAc])
pKa = 4.74, [C2H3O2^-]=0.012, [HOAc]= 0.004
You decide if your answer is correct