40 mL of 0.1 M HCl obviously will take exactly 40 mL of 0.1 M NaOH to arrive at the equivalence point. So a and b are before the eq pt and c is after the eq pt.
For a.
HCl + NaOH ==> NaCl + H2O.
millimols HCl = 40 x 0.1 = 4
mmols NaOH = 28 x 0.1 = 2.8
Difference [which is HCl which is (H^+)] = 1.2 millimols.
M = mmols/mL (mols/L) = 1.2/(28+40) = ?
Then pH = -log(H^+)
b. Done the same way.
c. After the equivalence point you have a solution of NaCl in extra base.
millimols NaOH at 48 mL = 48 x 0.1 = 4.8
mmols HCl at the eq pt = 40 x 0.1 = 4.0
mmols NaOH in excess is 0.8
(OH^-) = 0.8/(48+40) = ?
pOH = -log(OH^-)
pH + pOH = pKw = 14. You know pKw and pOH, solve for pH.
Calculate the pH during the titration of 40 ml of 0.1000 M HCl with 0.1000 M NaOH solution after the following additions of base:
A.) 28.00 ML
B.) 39.80 ML
C.) 48.00 ML
1 answer
1 answer