SME
Analytical Chemistry
Titrations
Use this for “0.4218 g of KHP needs 21.37 mL of NaOH to reach the phenolphthalein end point. What is the NaOH concentration?”
Diagram
The KHP is used up at 24.95 mL of NaOH, where the pH is 8.966: above 7, since the acid’s partner base is left.
After 24.95 mL of NaOH, the pH is 8.966.
This is an example. Type your own values to start a new problem.
C (M)
=
m (g)
÷
(
M (g/mol)
×
V (mL)
)
n
Moles of KHP
calculated · mmol
Formulas
n
=
m
M
×
1,000
2.5
=
0.5105
204.22
×
1,000
C
=
n
V
0.1002
=
2.5
24.95
Assumptions
•
KHP (potassium hydrogen phthalate, 204.22 g/mol) is a primary standard: pure, dried and weighed exactly.
•
KHP has one acidic hydrogen, so it reacts one to one with NaOH; the end point is the equivalence point.
•
The water the KHP is dissolved in (about 50 mL) changes no amount: only the moles count.
Step-by-step
WE KNOW
m = 0.5105 g, M = 204.22 g/mol, V = 24.95 mL
FIND
moles of KHP (n), concentration of NaOH (C)
Step 1 · Find moles of KHP (n)
Use n = m ÷ M × 1,000
Grams over grams per mole gives moles; × 1000 gives millimoles.
n
=
0.5105
204.22
×
1,000
n
=
0.0025
×
1,000
n
=
2.5
mmol
Step 2 · Find concentration of NaOH (C)
Use C = n ÷ V
At the end point the NaOH added matches the KHP mole for mole; millimoles per milliliter is molarity.
C
=
2.5
24.95
C
=
0.1002
M
Check
Put the numbers back in. Both sides should match.
2.5
=
0.5105
204.22
×
1,000
✓
0.1002
=
2.5
24.95
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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