SCH3U-R.H.KING ACADEMY
Titration - Neutralization Reactions
Name:
More Practice:
1. If 20.42 mL of Ba(OH)2 solution was used to titrate29.26 mL of 0.430 M HCl, what is the molarity of the
bariumhydroxide solution?[0.308 M]
#H x MA x VA = #OH x MB x VB4
MB = #H x MA x VA
#OH x VB
MB = 1 x 0.430M HCl x 0.02926 L
2 x 0.02042 L
MB= 0.0125818
0.04084
= 0.308 M of Ba(OH)2
2. What volume of 0.10 mol/L NaOH is needed to neutralize 25.0 mL of 0.15 mol/L H3PO4? [0.11L]
#H x MA x VA = #OH x MB x VB
VB = #H x MA x VA
#OH x MB
VB = 3 x 0.15M H3PO4 x 0.025 L
1 x 0.10 M
VB= 0.01125
0.1
= 0.11 L of NaOH
3. 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? [0.32]
#H x MA x VA = #OH x MB x VB
MA = #OH x MB x VB
#H x VA
MA = 2 x 0.1 M Ca(OH)2 x 0.04 L
1 x 0.025 L
MA= 0.008
0.025
= 0.32 M of HCl
4. A truck carrying sulfuric acid [ H2SO4] 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? [6.0M]
#H x MA x VA = #OH x MB x VB
MA = #OH x MB x VB
#H x VA
MA = 1 x 4.0 M NaOH x 0.06 L
2 x 0.02 L
MA= 0.24
0.04
= 6.0 M of H2SO4
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SCH3U-R.H.KING ACADEMY
Titration - Neutralization Reactions
Name:
5. What volume of 1.50 mol/L H2S will neutralize a solution containing 32.0 g NaOH? [0.267L]
H2S +2 NaOH --> SNa2 + 2H2O
# mols of H2S = 32.0g of NaOH X 1 mol of NaOH
40 of NaOH
X
1 mol of H2S
2 mol of NaOH
= 0.4 mols of H2S
Therefore, C=n/V
- V= n/C
- V= 0.4 mols/ 1.50M
- V= 0.27 L of H2S
6. A student finds that 37.80 mL of a 0.4052 M NaHCO3 solution is required to titrate a 20.00-mL sample of sulfuric
acid solution. What is the molarity of the acid? H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2 [0.3829M
#H x MA x VA = #OH x MB x VB
MA = #OH x MB x VB
#H x VA
MA = 1 x 37.80 mL x 0.4052 M
2 x 20.00mL
MA= 15.31656
40.00
=0.3829 M
7. What mass of Sr(OH)2 will be required to neutralize 19.54 mL of 0.00850 M HBr solution?[ 0.01010g]
Sr(OH)2S +2 HBr --> SrBr2 + 2H2O
grams of Sr(OH)2 = 0.01954 L of HBr X 0.00850 mol of HBr X 1 mol of Sr(OH)2 X 121.64g of Sr(OH)2
1L
2 mol of NaOH
1 mol of Sr(OH)2
= 0.01010g of Sr(OH)2
H2
SCH3U-R.H.KING ACADEMY
Titration - Neutralization Reactions
Name:
8. Calculate the pH of the solution that results when 100.0 mL of 0.03 M HCl (aq) reacts with 0.28 g of Mg(OH)2
pellets? Hint: Remember Limiting reagent. [pH= 12.78]
2 HCl (aq) + Mg (OH)2 --> MgCl2 + 2H2O
1) Limiting Reagent


Mg(OH)2
- n = m/mm
- n= o.28g/58.33g/mol
- 0.0048/1
= 0.0048 mols of Mg(OH)2
HCl
- C= n/V
- n= (C)(V)
- n= (0.03 M)(0.1 L)
- n= 0.003/2
- n=0.0015 mol
= HCl is the limiting factor
Therefore base in water Mg(OH)2 --> Mg 2+ + 2OHExcess= 0.0048-0.0015 = 0.0033 mols of Mg(OH)2
Therefore, 2(0.0033)= 0.0066 mols of OH


C of OH- = n/V
C= 0.0066/0.1000L
C= 0.066 M
Therefore, pOH
pOH = -log [OH]
- pOH = -log [0.066M]
= 1.18
pH + pOH = 14
- pH= 14-pOH
- pH=14-1.18
pH=12.82
This makes sense because there is an excess of Mg(OH)2 base, thus making the solution more basic. This
would result in a pH over 7.
9. What volume of a 2.0 M CaO solution (lime) would need to be placed into the lake 1,500,000,000 L (with known
molarity of 0.5M HCl) to raise the pH to 7. CaO + 2HCl  CaCl2 + H2O . Would this be realistic to do? Show
work![ V= 3.75x 10 8L]
#H x MA x VA = #OH x MB x VB
VB = #H x MA x VA
#OH x MB
VB = 1 x 0.5 M HCl x 1,500,000,000 mL
1 x2.0 M CaO
VB= 7.5 x 108
2.0M
= 3.75 x 10 8 L
S.D.= 3.8 x 10 8 L
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