Name___[KEY]______Date:______Period:______
[KEY]REVIEW Unit 10Test (Chp 17): Buffers Acid-Base Titrations[KEY]
Use these answers for questions 1 – 3
(A) a solution with a pH less than 7 that is not a buffer solution
(B) a buffer solution with a pH between 4 and 7
(C) a buffer solution with a pH between 7 and 10
(D) a solution with a pH greater than 7 that is not a buffer solution
(E) a solution with a pH of 7
A 1.A solution prepared to be initially 1 M in NaCl and 1 M in HCl.
D 2.A solution prepared to be initially 1 M in Na2CO3 and 1 M in CH3COONa.
B 3.A solution prepared to be initially 0.5 M in CH3COOH and 1 M in CH3COONa.
4.What change(s) will be caused by addition of a small amount of NaOH to a solution containing nitrite ions and nitrous acid? (hint: write two reactions – original rxn, then addition of NaOH)
original (buffer) solution:HNO2 + H2O NO2– + H3O+
Adding NaOH will decrease the HNO2 molecules (weak acid) and increase the NO2– ions (conjugate base)
because the acid-base rxn is…
acid-base rxn:HNO2 + OH– NO2– + H2O
___5.A solution of calcium hypochlorite, a common additive to swimming-pool water, is
(A) basic because of the hydrolysis of the OCl– ion
(B) basic because Ca(OH)2 is a weak and insoluble base
(C) neutral if the concentration is kept below 0.1 molar
(D) acidic because of the hydrolysis of the Ca2+ ions
(E) acidic because the acid HOCl is formed
___6.Mixtures that would be considered buffers include which of the following?
I. 0.50 M HC2H3O2 + 0.50 M NaC2H3O2
II. 0.50 M HNO3 + 0.50 MFe(NO3)2
III. 0.50 M NH3 + 0.50 MNH4Cl
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III (b/c weak with conjugate)
7.A 50.mL sample of 0.25MCa(OH)2 is titrated with a 0.50 MHCl solution. What is the [OH–](molarity) before any acid is added?
0.25 mol Ca(OH)2 x 2 mol OH– = 0.50 M OH–
1 mol Ca(OH)2
8.Ammonia, NH3, a common household cleaner, was titrated with HCl to determine its concentration. For 10.0 milliliters ofthe NH3, 20.0 millilitersof 0.100-molar HCl solution was required.
What was the concentration of NH3originally?
NH3 + HCl NH4+ + Cl–
0.0200 L HCl x 0.100 mol HCl x 1 mol NH3 = 0.00200 mol NH3
1 L HCl1 mol HCl
0.00200 mol NH3 = 0.200 M NH3
0.0100 L NH3 (no calculator needed if done step-by-step)
9.Write the net ionic equation for the reaction that occurs during the titration of hypobromous acid with barium hydroxide. What is ratio of acid to base in this titration?
Molecular 2 HOBr + Ba(OH)2 2 H2O + Ba(OBr)2
Net ionic HOBr + OH– H2O + OBr–
Acid:Base ratio is 2:1 based on overall balanced molecular equation.
It takes 2 HOBr’s to react with every 1 Ba(OH)2 .
Questions 10–13,
The graph below shows the titration curve that results when 100. mL of an unknown monoprotic acid HA is titrated with 0.100 M NaOH.
___10.Which of the following indicators is the best choice for this titration?
Indicator / pH Range of Color Change(A) Methyl orange 3.2 - 4.4
(B) Methyl red 4.8 - 6.0
(C) Bromothymol blue 6.1 - 7.6
(D) Phenolphthalein 8.2 - 10.0
(E) Alizarin 11.0 - 12.4
___11.What part of the curve corresponds to the optimum buffer action for the acid/conjugate ion pair?
(A) Point V
(B) Point X
(C) Point Z
(D) Along all of section WY
(E) Along all of section YZ
12. Based on the equivalence point shown in the graph, determine the concentration of the acid HA.
“monoprotic” acid means only one H so the rxn is:HA + NaOH H2O + NaA with a 1:1 ratio
Veq = 25.0 mL of 0.100 M NaOH according to the graph at point X.
0.0250 L NaOH x 0.100 mol NaOH x 1 mol HA = 0.00250 mol HA0.00250 mol HA = 0.0250 MHA
1 L NaOH 1 mol NaOH 0.100 L HA
13. Is HA a weak acid or a strong acid? List 3 reasons for your choice based on the graph.
weak acid because:
1)the initial pH is between 3 and 4 (not 1-2 like a typical strong acid)
2)the pH a equivalence is above 7 indicating the presence of a weak conjugate base is left
3)the pH rises gradually (rather remaining more level before equivalence like a strong acid)
Questions 14-15,
14.A pH curve for a certain titration is described in the following manner.
The graph of pH vs. mL of substance added lowers gradually at first, then drops steeply, followed by a leveling of pH at a low value.
Sketch the graph described on the axes below.
___15.Which of the following is the correct labeling of the substance titrated and the titrant (substance added) in question 14 above?
(A) strong base with strong acid
(B) weak base with strong acid (see 5 descriptions of the graph in #14 above)
(C) weak base with weak acid
(D) weak acid with weak base
(E) strong acid with weak base
___16.What is the end point of a titration?
(A) the moles of weak and conjugate are equal
(B) the moles of substance titrated and the moles of titrant are stoichiometrically equal
(C) the volume of titrant is equal to the volume of the substance titrated
(D) the indicator changes color
(E) the pH is equal to 7
Section II Free Response
CALCULATOR ALLOWED
CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS.
It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if you do not. Attention should be paid to significant figures.
C6H5COOH(aq) H+(aq) + C6H5COO–(aq)Ka = 6.46 x 10–5
- Benzoic acid, C6H5COOH, dissociates in water as shown in the equation above. A 25.0 mL sample of an aqueous solution of benzoic acid is titrated using standardized 0.150 M NaOH.
(a)After addition of 15.0 mL of the 0.150 M NaOH, the pH of the resulting solution is 4.37. Calculate each of the following.
(i)[H+] in the solution (1)
[H+] = 10^–4.37 = 4.3 x 10–5M(two sig figs due to two decimal places in pH)
(but 4.27 x 10–5 (3 sig figs) would typically be accepted)
(ii)The number of moles of NaOH added (1)
0.0150 L NaOH x 0.150 mol NaOH = 0.00225 mol NaOH
1 L NaOH
(iii)The number of moles of C6H5COO– in the solution (1)
C6H5COOH + OH– C6H5COO– + H2O
I ? mol 0.00225 mol 0
C –0.00225 –0.00225 +0.00225
E ? mol 0 0.00225 mol
(iv)[C6H5COOH] in the solution (2)
Total volume: V = (25.0 mL acid) + (15.0 mL NaOH)
V = 0.0400 L
[H+] = 4.3 x 10–5M from part (a)(i)
[C6H5COO–] = 0.00225 mol. = 0.0563 M
0.0400 L
Ka = [H+][C6H5COO–]
[C6H5COOH]
6.46 x 10–5 = (4.3 x 10–5)(0.0563)[C6H5COOH] = 0.037 M
[C6H5COOH]
(v)The number of moles of C6H5COOH in the solution (1)
0.0400 L x 0.037 mol C6H5COOH = 0.0015 mol C6H5COOH
1 L
In a different titration, a 0.7529 g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved in water and titrated with 0.150 M NaOH. The equivalence point is reached when 24.78 mL of the base solution is added.
(b)State whether the solution at the equivalence point is acidic, basic, or neutral. Explain your reasoning. (1)
At the equivalence point the solution is basic due to the presence of conjugate base (C6H5COO–)
that produces OH– in solution with water shown by the hydrolysis reaction:
C6H5COO– + H2O C6H5COOH + OH–
(c)Calculate each of the following.
(i)The moles of benzoic acid in the solid sample (1)
C6H5COOH + OH– C6H5COO– + H2O
0.02478 L NaOH x 0.150 mol NaOH x 1mol C6H5COOH = 0.00372 molC6H5COOH
1 L NaOH 1 mol NaOH
(ii)The mass, in grams, of benzoic acid in the solid sample (1)
molar mass of C6H5COOH = [7(12.01) + 6(1.008) + 2(16.00)]
= 122.12 g/mol
0.00372 mol C6H5COOH x 122.12 g C6H5COOH = 0.454 gC6H5COOH
1 mol C6H5COOH
(iii)The mass percentage of benzoic added in the solid sample (1)
mass % C6H5COOH = 0.454g C6H5COOH x 100 = 60.3%
0.7529 sample
H2NNH2(aq) + H2O(aq) H2NNH3+(aq) + OH–(aq)Kb = 3.0 x 10–6
- Hydrazine, a weak base, reacts with water according to the reaction represented above.
A sample of hydrazine is dissolved in water to produce 100.0 mL of a 0.10 M solution. This solution is titrated with 0.20 M HNO3.
(a)Write the balanced net-ionic equation for the reaction that occurs when the solution of HNO3 is added to thesolution of hydrazine. (1)
H2NNH2 + HCl H2NNH3+ + Cl–
H2NNH2 + H+ H2NNH3+
(b)Calculate the pH of the solution when 20.0 mL of the acid has been added. (3)
0.1000 L H2NNH2 x 0.10 mol H2NNH2 = 0.010 mol H2NNH2
1 LH2NNH2
0.0200 L HNO3 x 0.20 molHNO3 = 0.0040 mol HNO3
1 L HNO3
H2NNH2 + H+ H2NNH3+
I 0.010 mol 0.004 mol 0
C –0.0040 –0.0040 +0.0040
E 0.006 mol 0 0.004 mol
0.006 mol = 0.050 M 0.004 mol = 0.033 M
0.120 L 0.120 L
H2NNH2 + H2O H2NNH3+ + OH–
Kb = [H2NNH3+] [OH–]
[H2NNH2]
3.0 x 10–6 = (0.033)(x)
(0.050)
x = 4.5 x 10–6M OH–
pOH = –log [OH–]
pOH = –log (4.5 x 10–6) = 5.35
pH = 14 – pOH
pH = 14 – 5.35 = 8.65
(c)Calculate the volume of acid added at equivalence for the titration in part (a). (1)
0.100 L x 0.10 mol H2NNH2 x 1mol HNO3 x 1L HNO3 = 0.050L HNO3 = Veq
1 L1 mol H2NNH2 0.20 mol HNO3
(d)Calculate the pH at the equivalence point of the titration in part (a). (2)
H2NNH2 + HNO3 H2NNH3+ + NO3–
I0.010 mol 0.010 mol 0
C–0.010 –0.010 +0.010
E 0 0 0.010 mol0.10 mol = 0.067 M H2NNH3+
0.150 L
Now use an acid reaction in water and a Ka expression (not Kb b/c no longer any base left)
H2NNH3+ + H2O H2NNH2 + H3O+
Ka = [H3O+] [H2NNH2] Ka = Kw1.0 x 10–14 = 3.3 x 10–9= Ka
[H2NNH3+] Kb 3.0 x 10–6
H2NNH3+ + H2O H2NNH2 + H3O+
I 0.067 0 0
C –x +x +x
E0.067–x x x
≈ 0.067
Ka = [H3O+] [H2NNH2]3.3 x 10–9 = x2 x = [H3O+] = 1.5 x 10–5M
[H2NNH3+] 0.067
pH = –log [H3O+]
pH = –log (1.5 x 10–5) = 4.8
(e)The pKa values for several indicators are given below. Which of the indicators listed is most suitable for this titration? Justify your answer. (1)
Indicator / pKaThymol blue / 2
Methyl orange / 4
Phenolphthalein / 9
The pH at equivalence is acidic (4.8), so the best indicator is methyl orange for which the value of pKa is closest to the pH at equivalence.
- A 1.22 g sample of a pure monoprotic acid, HA, was dissolved in distilled water. The HA solution was thentitrated with 0.250 M KOH. The pH was measured throughout the titration, and the equivalence point wasreached when 40.0 mL of the KOH solution had been added. The data from the titration are recorded in thetable below.
Volume of 0.250 M KOH Added (mL) / pH of Titrated Solution
0.00 / ?
10.0 / 3.72
20.0 / 4.20
30.0 / ?
40.0 / 8.62
50.0 / 12.40
(a)Explain how the data in the table above provide evidence that HA is a weak acid rather than a strong acid. (1)
pH at equivalence is > 7 due to conjugate base reaction with H2O producing OH–
(b)Write the balanced net-ionic equation for the reaction that occurs when the solution of KOH is added to thesolution of HA. (1)
HA + OH– A– + H2O
(c)Calculate the number of moles of HA that were titrated. (1)
0.0400 L NaOH x 0.250 mol NaOH x 1 mol HA = 0.0100 mol HA
1 L NaOH 1 mol NaOH
(d)Calculate the molar mass of HA . (1)
1.22 g HA = 122 g/mol
0.0100 mo HA
The equation for the dissociation reaction of HA in water is shown below.
HA(aq) + H2O(l) H3O+(aq) + A−(aq)Ka = 6.3 × 10−5
(e)Assume that the initial concentration of the HA solution (before any KOH solution was added) is 0.200 M . Determine the pH of the initial HA solution. (2)
HA(aq) + H2O(l) H3O+(aq) + A−(aq)
I 0.200M 0M 0M
C –x +x +x
E0.200–x x x
≈ 0.200 M b/c K < 1
Ka = [H3O+][A–]
[HA]
6.3 x 10–5 = x2
(0.200)
x = 3.5 x 10–3M H3O+
pH = –log (3.5 x 10–3) = 2.45
(f)Calculate the value of [H3O+] in the solution after 30.0 mL of 0.250 M KOH solution is added to 50.0 mL of 0.200 M HA solution. (3)
0.0500 L HA x 0.10 mol HA = 0.0100 mol HA
1 LHA
0.0300 L KOH x 0.250 mol KOH = 0.00750 mol KOH
1 L KOH
HA + OH– A– + H2O
I 0.0100 mol 0.00750 mol 0 mol
C –0.00750 –0.00750 +0.00750
E 0.00250 mol 0 mol 0.00750 mol
0.00250 mol = 0.0313 M HA 0.00750 mol = 0.0938 M A–
0.0800 L 0.0800 L
HA(aq) + H2O(l) H3O+(aq) + A−(aq)
I 0.0313 M 0 M 0.0938 M
C –x +x +x
E0.0313–x x 0.938 + x
≈ 0.0313M b/c K < 1≈ 0.0938 M b/c K < 1
6.3 x 10–5 = (x)(0.0938)
(0.0313)
x = 2.1 x 10–5M H3O+
- An approximately 0.1–molar solution of NaOH is to be standardized by titration with a primary standard. Assume that the following materials are available.
Clean, dry 50 mL buret250 mL Erlenmeyer flask
25 mL pipet Wash bottle filled with distilled water
Analytical balance10 mL graduated cylinder
Phenolphthalein indicator100 mL volumetric flask
Potassium hydrogen phthalate, KHP, a pure solid monoprotic acid (a primary standard)
(a)Briefly describe a procedure and indicate the measurements to be made, using the materials listed above, to standardize the NaOH solution. (You may not use all of the materials) (3)
1)Record the exact mass of the 100 mL volumetric flask with the balance.
2)Record the exact mass of a sample of KHP in the volumetric flask.
3)Add distilled water to dissolve the solid KHP in the volumetric flask.
Fill to the mark, cap it, and mix it gently to ensure homogeneity.
4)Pipet 25 mL of NaOH solution into the 250 mL Erlenmeyer flask.
5)Add a few drops of phenolphthalein indicator to the flask.
6)Rinse the buret with the KHP solution, then fill it.
7)Run out some KHP titrant into a waste beaker to remove air bubbles.
8)Recordthe initial volume of KHPsolution in the buret.
9)While mixing, titrate the NaOHsolution with the KHP solution until it remains pink.
10)Recordthe final volume of KHP solution in the buret.
(b)Describe at least three calculations necessary to determine the concentration of the newly standardized NaOH solution. (2)
grams of KHP = (mass of solid KHP in flask) – (mass of empty flask)
molarity of KHP = (grams of KHP) ÷ (molar mass of KHP)
volume of KHP = (final volume in buret) – (initial volume in buret)
moles of KHP = (volume of KHP) x (molarity of KHP)
moles of NaOH = moles of KHP (1:1 mole ratio)
molarity of NaOH = (moles NaOH) ÷ (volume of NaOH)
(c)After the NaOH solution has been standardized, it is used to titrate 10.0 mL of a weak monoprotic acid, HX. The equivalence point is reached when 25.0 mL of NaOH solution has been added. On the graph provided below, sketch the titration curve, showing the pH changes that occur as the volume of NaOH solution added increases from 0 to 35.0 mL. Clearly label the equivalence point on the curve. (2)
(d)Explain what effect each of the following changes in the titration procedure listed in part (c) would have on the calculated concentration of HX . Justify your answers.
(i)The buret used to titrated the HX solution was rinsed with distilled water before it was filled with standardized NaOH solution. (1)
When standardized NaOH is added to the buret, it will be diluted by the left over distilled water in the buret from rinsing.
It will take more volume of (diluted) NaOH to titrate the acid, HX, to equivalence which will make the moles of NaOH consumed seem greater.
The moles of HX reacted will also seem greater.
Therefore, the calculated concentration of HX will betoo large.
(ii)A 20.0 mL sample of HX solution is titrated rather than a 10.0 mL sample. (1)
A 20.0 mL sample of HX would take more volume of NaOH to titrate to equivalence.
More moles of NaOH will be consumed and more moles of HX will be reacted.
However, more moles of HX reacted divided by the greater volume if HX titrated will equal the same concentration (molarity) of HX.
Therefore, there will be no effect on the calculated concentration of HX.
(e)The graph at the right shows the results obtained by
titrating a different weak acid, H2Y, with the
standardized NaOH solution.
Identify the negative ionthat is presentin the
highest concentration at the pointin the titration
represented by the letter A on the curve. (1)
Y2–
ANSWER KEY
1.2006B#1(modified slightly)
2.2003#1(c)(d)(e)(modified version with different substances and/or values.)
3.2012#1
3.1998D(modified version of “ACID BASE 1998D Required”)
1