196
Stolchlometry III: Calculations Based on Concentrations of Solutions
PROBLEM:
Titration of a 0.7865 g sample of pure potassium hydrogen phthalate requires 35.73
ml of a NaOH solution. Calculate the molarity of the NaOH solution. This acidbase reaction is
NaOH + KHC 8 H 4 O 4 -H> NaKC 8 H 4 O 4 + H 2 O
OH- + HC 8 H 4 O- 4 -» C 8 H 4 OJ- + H 2 O
SOLUTION:
1. From the given grams of KHC 8 H 4 O 4 , calculate the moles:
moles of KHC 8 H 4 0 4 given =
-
g KHC 8 H 4 O 4 = 3 ^ x 1Q _
2041 g KHC 8 H 4 0 4
mole KHC 8 H 4 O 4
2. The chemical equation shows 1 mole of NaOH needed per mole of
KHC 8 H 4 O 4 at the endpoint, so
moles of NaOH used = ( , * 7°!!,^°" ) (3.854 x 10^ moles KHC 8 H 4 O 4 )
\ 1 mole K.HC 8 H 4 O 4 /
= 3.854 x 10J moles NaOH
3. The moles of NaOH used were dissolved in 35.73 ml of solution. Therefore,
the molarity of the solution is
, .
3.854 x 1(T
3 moles NaOH
„ „„„ moles NaOH
molarity
=
-
=
0. 1079
-
0.03573 liters
liter
You have "standardized" the solution and found it to be 0.1079 M NaOH.
PROBLEM:
A solution of H 2 SO 4 is prepared by diluting commercial concentrated acid. A
volume of 42.67 ml of this solution is required to titrate exactly 50.00 ml of the
NaOH solution standardized in the preceding problem. Calculate the molarity of
the H 2 SO 4 solution. The titration reaction is
2NaOH + H 2 SO 4 -> Na 2 SO 4 + 2H 2 O
OH- + H
+ ^ H 2 O
SOLUTION:
1. From the given volume and concentration of NaOH solution, calculate the
moles of NaOH given (used):
moles of NaOH used = (0.05000 liters) ((U079
moles NaOH\
\
liter
/
= 5.395 x 10~
3 moles NaOH
2. The chemical equation shows that 1 mole of H 2 SO 4 is used per 2 moles of
NaOH at the endpoint, so the number of moles of H 2 SO 4 used is