The secret to these problems is to know where you are on the titration curve; i.e., beginning (0 mL), equilvalence point, before the eq pt or after the eq pt.
You can calculate the eq pt easily.
mL acid x M acid = mL base x M base (that is always true for MONOPROTIC acids titrated with monobasic bases).
41.2 x 0.230 = mL base x 0.220
mL base = estimated 43 mL
Therefore, you know 11.9 mL is between the beginning and the eq. pt and those pHs are determined by the Henderson-Hasselbalch equation.
pH = pKa + log(base)/(acid)
where (base) is the (F^-) and (acid) is the (HF).
mols HF initially = M x L = 0.230 x 0.0412 = 0.00948
mol NaOH added = M x L = 0.220 x 0.0119 = 0.00262
...........HF + NaOH ==> NaF + H2O
I.....0.00948....0........0......0
added..........0.00262..........
C...-0.00262..-0.00262...0.00262......
E.....0.00686...0........0.00262
Ka = 3.5E-4; pKa = -log Ka = 3.46
You can use a shortcut and use mols instead of concn which makes it
pH = 3.46 + log((0.00262/0.00686) = ?
Consider the titration of 41.2ml of 0.230 M HF with 0.220 M NaOH . Calculate the ph after the addition of 11.9ml of sodium hydroxide.(HF: Ka = 3.50 x 10-4)
Please, I need step by step on on how to solve this type of problem.
Thank you very much!
2 answers
Just a quick note to say that I rounded my numbers above so you should go through the entire calculation and change numbers that I rounded more or less than I should have done. My answer is close to the correct value but it may not be exact.
1 answer