SME
Analytical Chemistry · Problem type
Back titration
Use this for “An antacid tablet dissolves in 40.00 mL of 0.2000 M HCl, and the excess acid takes 12.35 mL of 0.1500 M NaOH. How many mg of CaCO₃ did it hold?”
Diagram
3.18 + 1.82 = 5
This is an example. Type your own values to start a new problem.
nreacted (mmol)
=
CHCl (M)
×
VHCl (mL)
−
CNaOH (M)
×
VNaOH (mL)
mCaCO₃ (mg)
=
nreacted (mmol)
÷
r
×
100.09
g/mol
nadded
Moles of HCl added
calculated · mmol
nleft
Moles of HCl left over
calculated · mmol
nCaCO₃
Moles of CaCO₃
calculated · mmol
Formulas
nadded = CHCl × VHCl
5 = 0.1 × 50
nleft = CNaOH × VNaOH
1.82 = 0.1 × 18.2
nreacted = nadded − nleft
3.18 = 5 − 1.82
nCaCO₃
=
nreacted
r
1.59
=
3.18
2
mCaCO₃ = nCaCO₃ × 100.09
159.1 = 1.59 × 100.09
Assumptions
•
A measured excess of HCl reacts with all the analyte; NaOH then titrates the acid left over, one to one.
•
The acid that reacted is the acid added less the acid left over.
•
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂: the mole ratio is 2 HCl per CaCO₃ (100.09 g/mol).
Step-by-step
WE KNOW
CHCl = 0.1 M, VHCl = 50 mL, CNaOH = 0.1 M, VNaOH = 18.2 mL, r = 2
FIND
moles of HCl added (nadded), moles of HCl left over (nleft), moles of HCl that reacted (nreacted), moles of CaCO₃ (nCaCO₃), mass of CaCO₃ (mCaCO₃)
Step 1 · Find moles of HCl added (nadded)
Use nadded = CHCl × VHCl
Molarity times milliliters gives millimoles of HCl added.
nadded = 0.1 × 50
nadded = 5 mmol
Step 2 · Find moles of HCl left over (nleft)
Use nleft = CNaOH × VNaOH
NaOH takes up the leftover HCl one to one, so its millimoles are the HCl left over.
nleft = 0.1 × 18.2
nleft = 1.82 mmol
Step 3 · Find moles of HCl that reacted (nreacted)
Use nreacted = nadded − nleft
The acid the CaCO₃ used up is what was added less what was left for the NaOH.
nreacted = 5 − 1.82
nreacted = 3.18 mmol
Step 4 · Find moles of CaCO₃ (nCaCO₃)
Use nCaCO₃ = nreacted ÷ r
Each
mole
of
CaCO₃
used
r
moles
of
HCl,
so
divide
by
the
ratio.
nCaCO₃
=
3.18
2
nCaCO₃ = 1.59 mmol
Step 5 · Find mass of CaCO₃ (mCaCO₃)
Use mCaCO₃ = nCaCO₃ × 100.09
CaCO₃ is 100.09 g/mol, which is 100.09 mg per millimole.
mCaCO₃ = 1.59 × 100.09
mCaCO₃ = 159.1 mg
Check
Put the numbers back in. Both sides should match.
5 = 0.1 × 50 ✓
1.82 = 0.1 × 18.2 ✓
3.18 = 5 − 1.82 ✓
1.59
=
3.18
2
✓
159.1 = 1.59 × 100.09 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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