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Lecture 33
Acids and Bases V
Tutorial
1) A 65 mL sample of 0.35 M NH3 is titrated with 0.25 M HCl. Kb for ammonia is
1.8 x 10-5.
a. Find the initial pH of the 0.35 M NH3 solution.
NH 3( aq )
I
C
E
+
0.35
-x
0.35 - x
H 2O(l )
~
→ NH 4 + ( aq ) +
0
+ x
x
OH − ( aq )
0
+ x
x
[NH 4 + ][OH − ]
( x)( x)
Kb =
=
[NH 3 ]
0.35 - x
assume 0.35 - x = 0.35
x2
0.35
2
x = (0.35)(1.8 ×10−5 )
1.8 × 10−5 =
x = (0.35)(1.8 × 10−5 )
x = [OH + ] = 2.5 ×10−3 M
pOH = − log[OH + ] = − log(2.5 ×10−3 M ) = 2.6
pH = 14 − pOH = 14 − 2.6 = 11.4
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b. Find the pH of the solution after 31 mL of 0.25 M HCl have been added.
Step 1) Find the initail number of moles of NH 3 and H +
0.35 mol NH 3
0.25 mol H +
0.065 L×
= 0.023 mol NH 3
0.031L ×
=0.0078 mol H +
1L
1L
+
Step 2) Find the number of moles of NH 3 , and NH 4 after reaction
H+
+
NH 3
→
NH 4 +
I
0.0078 mol
0.023 mol
0 mol
C
-0.0078 mol
E
0 mol
-0.0078 mol
0.015 mol
+0.0078 mol
0.0078 mol
Step 3) Find concentrations
New Volume = 0.065 L + 0.031 L = 0.096 L
0.015 mol
= 0.16 M NH3
0.096 L
Step 4) Find pOH and pH
MolarityNH3 =
pOH = pK b + log
MolarityNH + =
4
0.0078 mol
= 0.081 M NH 4 +
0.096 L
[NH 4 + ]
(0.081M )
= − log(1.8 × 10−5 ) + log
= 4.4
[NH 3 ]
(0.16M )
pH = 14 − pOH = 14 − 4.4 = 9.6
c. Find the pH at the equivalence point.
Step 1) Find the number of moles of NH 3 and H + that have reacted
0.35 mol NH 3
= 0.023 mol NH 3 = mol H +
1L
Step 2) Find the volume of titrant used
1L
0.023 mol H + ×
= 0.092 L of 0.25 M HCl solution
0.25 mol H +
Step 3) Find concentration of conjugate acid
0.065 L×
New Volume = 0.092 L + 0.065 L = 0.157 L
[NH 4 + ] =
0.023 mol NH 4 +
= 0.15M NH 4 +
0.157 L
Step 4) Find K a for NH 4 +
Ka =
K w 1.0 ×10−14
=
= 5.6 ×10−10
−5
K b 1.8 ×10
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Step 5) Find [H + ]
NH 4 +
Ka =
→
NH3
I
C
0.15 M
-x
0M
+x
E
0.15 - x
x
H+
+
0M
+x
x
[NH 3 ][H + ]
(x)(x)
x2
=
≈
[NH 4 + ]
0.15 - x 0.15
x = [H + ] = K a (0.15) = (5.6 ×10−10 )(0.15) = 9.2 ×10−6 M
Step 6) Find pH
pH = - log[H + ] = -log(9.2 × 10−6 ) = 5.0
d. Find the pH of the solution after 151 mL of 0.25 M HCl have been added.
Step 1) Find the initail number of moles of NH 3 and H +
0.35 mol NH 3
=0.023 mol NH 3
1L
0.25 mol H +
0.151 L×
=0.038 mol H +
1L
Step 2) Find the number of moles of H + , NH 3 , and NH 4 + after reaction
0.065 L×
H+
I
C
E
+
0.038 mol
-0.023 mol
0.015 mol
NH 3
→
0.023 mol
-0.023 mol
0
NH 4 +
0 mol
+0.023 mol
0.023 mol
Step 3) Find [H + ]
New Volume = 0.065 L + 0.151 L = 0.216 L
MolarityH+ =
0.015 mol
= 0.069 M H +
0.216 L
Step 4) Find pH
pH = - log[H + ] = − log(0.069) = 1.2
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e. Draw a titration curve for this titration.
f. Which of the following indicators should be used to signal the endpoint of
this titration: methyl red (pKa = 5.5), litimus (pKa = 7.0), or
phenolphthalein (pKa = 8.7)? Justify your answer.
Methyl Red would be the best indicator, as pKa(methyl red) is very close to the pH of the
solution at its endpoint.
2) A beaker containing 125 mL of 0.120 M HOCl is titrated using 0.250 M NaOH.
Ka for HOCl is 3.5 x 10-8.
a. What volume of NaOH must be delivered to reach the equivalence point?
0.120 mol HOCl
= 0.0150mol HOCl
1L
1L
1000 mL
0.0150 mol NaOH×
×
=60.0 mL
0.250 mol NaOH
1L
0.125 L×
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b. What is the pH at the equivalence point?
Find concentration of conjugate base
New Volume = 0.060 L + 0.125 L = 0.185 L
0.0150 mol OCl= 0.0811M OCl−
[OCl ] =
0.185 L
−
Find K b for OCl
−
Kb =
K w 1.0 ×10−14
=
= 2.9 × 10−7
K a 3.5 × 10−8
Find [OH − ]
OCl−
I
C
E
Kb =
0.0811M
-x
0.0811 - x
+
H 2O
→
~
~
~
HOCl
+
0M
+x
x
OH −
0M
+x
x
[HOCl][OH − ]
(x)(x)
x2
=
≈
[OCl− ]
0.0811 - x 0.0811
x = K b (0.0811) = (2.9 × 10−7 )(0.0811) = 1.5 ×10−4 M
[OH − ] = x = 1.5 × 10−4 M
pOH = -log[OH − ] = -log(1.5 ×10−4 M ) = 3.8
pH = 14 - pOH = 10.2
c. Why is the solution basic and the equivalence point?
HOCl is a weak acid, so it conjugate base, OCl-, is relatively strong. At the equivalence
point equal numbers of moles of HOCl and OH- have reacted. The only species
remaining in solution that can change the pH is OCl-, which is basic. OCl- accepts
protons from water molecules, thereby increasing the concentration of OH- in the solution
according to the reaction below.
OCl-(aq)+ H2O(l) Æ HOCl(aq) + OH-(aq)
The pH rises into the basic range as [OH-] rises.
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3) A 1.02 g sample containing solid HF and LiCl is dissolved in distilled water. The
solution is then titrated with a standardized 0.250 M NaOH solution. The
equivalence point is reached when 35.72 mL of 0.250 M NaOH is delivered.
a. Find the number of moles of solid HF in the original sample.
mol HF = mol NaOH at equivalence point
0.25 mol NaOH
=0.0089 mol NaOH
1L
nHF = 0.0089 mol HF
0.03572 L ×
b. Find the mass of solid HF in the original sample.
0.0089 mol HF×
20.01g HF
=0.18g HF
1mol HF
c. Find the mass % of HF is the solid sample.
mass % =
mass HF
0.18g
× 100 =
× 100 = 18% HF by mass
1.02g
total mass
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