NaOH + HCl
NaOH
HCl
Volume
used (mL)
Molarity (M)
NaOH + H2SO4
NaOH H2SO4
10
34.3
10
17.1
0.6
0.175
0.6
0.175
1. Neutralization
2. NaOH + HCl  H2O + NaCl
2NaOH + H2SO4  2H2O + Na2SO4
3. Twice as much HCl was required. Because it
takes twice as much HCl (one H) as H2SO4
(two Hs) to neutralize the same amount of
NaOH
4. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq)
# L H2SO4=
0.010 L NaOHx0.60 mol NaOH 1 mol H2SO4
L H2SO4
x
x
L NaOH 2 mol NaOH 0.175 mol H2SO
= 0.01714 L = 17.1 mL
5. #H x MA x VA = #OH x MB x VB
Titration problems
1. What volume of 0.10 mol/L NaOH is needed
to neutralize 25.0 mL of 0.15 mol/L H3PO4?
2. 25.0 mL of HCl(aq) was neutralized by 40.0
mL of 0.10 mol/L Ca(OH)2 solution. What
was the concentration of HCl?
3. A truck carrying sulfuric acid is in an accident.
A laboratory analyzes a sample of the spilled
acid and finds that 20 mL of acid is neutralized by 60 mL of 4.0 mol/L NaOH solution.
What is the concentration of the acid?
4. What volume of 1.50 mol/L H2S will neutralize a solution containing 32.0 g NaOH?
Titration problems
1. (3)(0.15 M)(0.0250 L) = (1)(0.10 M)(VB)
VB= (3)(0.15 M)(0.0250 L) / (1)(0.10 M) = 0.11 L
2. (1)(MA)(0.0250 L) = (2)(0.10 M)(0.040 L)
MA= (2)(0.10 M)(0.040 L) / (1)(0.0250 L) = 0.32 M
3. Sulfuric acid = H2SO4
(2)(MA)(0.020 L) = (1)(4.0 mol/L)(0.060 L)
MA = (1)(4.0 M)(0.060 L) / (2)(0.020 L) = 6.0 M
4. mol NaOH = 32.0 g x 1 mol/40.00 g = 0.800
(2)(1.50 mol/L)(VA) = (1)(0.800 mol)
VA= (1)(0.800 mol) / (2)(1.50 mol/L) = 0.267 L
Titration summary
• For titrations we use the formula:
#H x MA x VA = #OH x MB x VB
Or NA x VA =
NB x VB
• NA is the combination of MA and #H
• N is also known as normality
• You can think of it a neutralizing power
• We will stick with the first equation, you do not
have to know N or what it stands for
• This is a simplification of stoichiometry. We
could get the same answer by working with
moles (n = MV) and by using the balanced
chemical equation
Titration showdown
• Titration competition (best
with a burette): find the
concentration of an H2SO4
solution – it could be
anywhere from 0-18 mol/L
• You will use the NaOH that you prepared two
weeks ago and your vast knowledge of titration
procedures and formulas.
• The wining team, is the team closest to the
correct value.
• The only restriction is : don’t put base in the
burette – only acid.
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Sources of error
NaOH was not exactly 0.10 mol/L
Calculation errors (e.g. converting mL to L)
Not rinsing and drying the beaker for the acid
Over titrating (ideally, 2 titrations should be
done – one to get a rough estimate, and one to
get the exact value).
Acid or base left on the side of the flask or on
the tip of the burette (for greater precision,
water is used to rinse the tip of the burette).
Errors reading volumes.
Using pipette (10 mL is measured from 0 mL to
10 mL, not from 10 mL to empty)
Using 10 mL of base vs. 25-50 mL
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