Titration Problems - Ap Chemistry 3g421a

Titration Problems 1973 A sample of 40.0 milliliters of a 0.100 molar HC 2H3O2 solution is titrated with a 0.150 molar NaOH solution. K a for acetic acid = 1.8x10-5

= MbVb = (0.0274 L)(0.135 M) = 3.70x10-3 mol HA molec .wt . =

mass HA

=

0 . 682 g 3 . 70 ∞ 10

mol HA

−3

= 184 g/mol

(b) (c) HA + OH- --> A- + H2O (a) What volume of NaOH is used in the titration in initial: 0.00370 mol order to reach the equivalence point? added: (0.0106L)(0.135M) = 0.00143 mole (b) What is the molar concentration of C2H3O2- at the remaining: (0.00370 - 0.00143) = 0.00227 mol equivalence point? (c) What is the pH of the solution at the equivalence (d) pH = -log[H3O+]; [H3O+] = 10-pH = 10-5.65 point? = 2.2x10-6M + − −6 Answer: [ H 3 O ][ A ] ( 2 . 2 ∞ 10 ) ( 0 . 00143 / v ) = = K (a) MaVa=MbVb a [ HA ] ( 0 . 00227 / v ) (e) (0.100M)(40.0 mL) = (0.150M)(Vb) = 1.4x10-6 Vb = 26.7 mL 1979 B (b) acetate ion is a weak base with A solution of hydrochloric acid has a density of 1.15 grams per milliliter and is 30.0% by weight HCl. Kb=Kw/Ka = 1.0×10-14/1.8×10-5 = 5.6×10-10 4 . 00 mmol − = 0 . 0600 M (a) What is the molarity of this solution of HCl? [ CH 3 COO ] o = ( 40 . 0 mL + 26 . 7 mL ) (b) What volume of this solution should be taken in order to prepare 5.0 liters of 0.20 molar [CH3COO ]eq = 0.600M -X hydrochloric acid by dilution with water? [OH ] = [CH3COOH] = X 2 (c) In order to obtain a precise concentration, the 0.20 X − 10 −5 = 5 . 6 ∞ 10 ; X = 9 . 66 ∞ 10 M molar hydrochloric acid is standardized against ( 0 . 0600 − X ) pure HgO (molecular weight = 216.59) by 0.0600M-9.66×10-5M = 0.0599M [CH3COO-]eq titrating the OH- produced according to the − 14 following quantitative reaction. Kw 1 . 0 ∞ 10 + − 10 = = 1 . 04 ∞ 10 [H ] = M − −5 HgO(s) + 4 I- + H2O --> HgI42- + 2 OH[ OH ] 9 . 66 ∞ 10 (c) In a typical experiment 0.7147 grams of HgO pH = -log [H+] = -log(1.04×10-10) = 9.98 required 31.67 milliliters of the hydrochloric acid 1978 A solution for titration. Based on these data what is A 0.682 gram sample of an unknown weak monoprotic organic the molarity of the HCl solution expressed to four acid, HA was dissolved in sufficient water to make 50 milliliters significant figures. of solution and was titrated with a 0.135 molar NaOH solution. After the addition of 10.6 milliliters of base, a pH of 5.65 was Answer: recorded. The equivalence point (end point) was reached after the addition of 27.4 milliliters of the 0.135 molar NaOH.

1 . 15 g 1 mL

∞

1000 mL 1L

∞

30 . 0 gHCl 100 g

∞

mol

1 mol 35 . 5 g

= 9 .5 M

(a) Calculate the number of moles of acid in the (a) original sample. (b) MfVf = MiVi (0.20M)(5.0L) = (9.5M)(V) (b) Calculate the molecular weight of the acid HA. V = 0.11 L (c) Calculate the number of moles of unreacted HA 0 . 7147 g remaining in solution when the pH was 5.65. = 0 . 003300 mol HgO (d) Calculate the [H3O+] at pH = 5.65 (c) 216 . 59 g mol (e) Calculate the value of the ionization constant, Ka, mol OH- prod. = 2 (mol HgO) = 0.006600 mol of the acid HA. mol HCl req. = mol OH- prod. = 0.006600 mol Answer: 0 . 006600 mol = 0 . 2084 M M HCl = (a) at equivalence point, moles HA = moles NaOH 0 . 03167 L

1982 D A solution of barium hydroxide is titrated with 0.1-M sulfuric acid and the electrical conductivity of the solution is measured as the titration proceeds. The data obtained are plotted on the graph below.

The molecular weight of a monoprotic acid HX was to be determined. A sample of 15.126 grams of HX was dissolved in distilled water and the volume brought to exactly 250.00 millilitres in a volumetric flask. Several 50.00 millilitre portions of this solution were titrated against NaOH solution, requiring an average of 38.21 millilitres of NaOH. The NaOH solution was standardized against oxalic acid dihydrate, H2C2O4.2H2O (molecular weight: 126.066 gram mol-1). The volume of NaOH solution required to neutralize 1.2596 grams of oxalic acid dihydrate was 41.24 millilitres.

Co n d u ctiv ity ,

λ

10

20

30

40

50

60

70

80

(a) Calculate the molarity of the NaOH solution. (b) Calculate the number of moles of HX in a 50.00 millilitre portion used for titration. (c) Calculate the molecular weight of HX. (d) Discuss the effect of the calculated molecular weight of HX if the sample of oxalic acid dihydrate contained a nonacidic impurity. Answer:

Millilitres of 0.1-M H2SO4

(a) For the reaction that occurs during the titration described above, write a balanced net ionic equation. (b) Explain why the conductivity decreases, es through a minimum, and then increases as the volume of H2SO4 added to the barium hydroxide is increased. (c) Calculate the number of moles of barium hydroxide originally present in the solution that is titrated. (d) Explain why the conductivity does not fall to zero at the equivalence point of this titration. Answer: (a) Ba2+ + 2 OH- + 2 H+ + SO42- --> BaSO4(s) + 2 H2O (b) The initial conductivity is high because of the presence of Ba2+ and OH- ions. The conductivity decreases because Ba2+ forms insoluble BaSO4 with the addition of SO42-. The conductivity also decreases because OH- combines with the addition of H+ ions by forming H2O. Beyond the equivalence point conductivity increases as H+ and SO42- ions are added. (c) # mol Ba(OH)2 = # mol H2SO4 =0.1M × 0.04L = 0.004 mol

1.2596 g g

mol = (a) mol H2C2O4.2H2O = 126.066 9.9916x10-3 mol H2C2O4 + 2 NaOH --> Na2C2O4 + 2 H2O

9 . 9916 ∞ 10

−3

2 molNaOH

mol ∞

= 1 . 9983 ∞ 10

−2

mol

1 molH 2 C 2 O 4 M NaOH =

−2 1. 9983 ∞ 10 mol

= 0 . 4846 M

0 . 04124 L

(b) mol HX = mol NaOH 0.03821 L x 0.4846 M = 0.01852 mol HX 0 . 01852 mol

(c)

∞ 250 . 00 mL = 0 . 09260 mol HX

50 . 00 mL MW

15 . 126 g

=

= 163 . 3

0 . 09260 mol

g mol

(d) The calculated molecular weight is smaller than true value, because: measured g H2C2O4 is larger than true value, calculated mol H2C2O4 is larger than true value, calculated mol NaOH is larger than true value, calculated M NaOH is larger than true value calculated mol HX is larger than true value, therefore, MW

g HX

=

mol HX

( true value

( calculated

)

, and too large

)

(d) BaSO4(s) dissociates slightly to form Ba2+ and 1984 C SO42-, while the water ionizes slightly to form H + Discuss the roles of indicators in the titration of acids and bases. and OH-. Explain the basis of their operation and the factors to be 1983 B

considered in selecting an appropriate indicator for a particular titration.

28 . 35 mLNaOH 0 . 2764 mol 1 molHNO 3 Answer: ∞ ∞ = 1L 1 molNaOH An indicator signals the end point of a titration by (b) 25 . 00 mLHNO 3 changing color. = 0.3134M HNO3 An indicator is a weak acid or weak base where the MfVf=MiVi;(0.3134M)(500mL) = (M)(10.00mL) acid form and basic form of the indicators are of different colors. M = 15.67M HNO3 g L HNO 3 An indicator changes color when the pH of the ∞ 100 % % HNO 3 in conc . sol ’n = solution equals the pKa of the indicator. In selecting an g L sol ’ n (c) indicator, the pH at which the indicator changes color grams HNO3 in 1 L conc. solÆn = should be equal to (or bracket) the pH of the solution 15.67 molHNO 3 63 . 02 g g at the equivalence point. ∞ = 987 . 5 L 1L 1 mol For example, when a strong acid is titrated with a strong base, the pH at the equivalence point is 7, so we grams sol’n in 1 L conc. Sol’n would choose an indicator that changes color at a pH = 1 . 42 g sol’n 1000 mL g ∞ = 1420 7. {Many other examples possible.} L 1 mL 1L

1987 B The percentage by weight of nitric acid, HNO 3, in a sample of concentrated nitric acid is to be determined.

(a) Initially a NaOH solution was standardized by titration with a sample of potassium hydrogen phthalate, KHC8H4O4, a monoprotic acid often used as a primary standard. A sample of pure KHC 8H4O4 weighing 1.518 grams was dissolved in water and titrated with the NaOH solution. To reach the equivalence point, 26.90 millilitres of base was required. Calculate the molarity of the NaOH solution. (Molecular weight: KHC8H4O4 = 204.2) (b) A 10.00 millilitre sample of the concentrated nitric acid was diluted with water to a total volume of 500.00 millilitres. Then 25.00 millilitres of the diluted acid solution was titrated with the standardized NaOH solution prepared in part (a). The equivalence point was reached after 28.35 millilitres of the base had been added. Calculate the molarity of the concentrated nitric acid. (c) The density of the concentrated nitric acid used in this experiment was determined to be 1.42 grams per millilitre. Determine the percentage by weight of HNO3 in the original sample of concentrated nitric acid. Answer: 1 mol

1 . 518 g ∞

(a)

= 7 . 434 ∞ 10

−3

mol acid

204 . 2 g

= mol NaOH required to neut. 7 . 434 ∞ 10

−3

0 . 02690 L

mol

= 0 . 2764 M NaOH

[ HA ] =

K =

( 0 . 0152 3 L )( 0 . 2211 mol L ) = 0 . 04810 M 0 . 0700 0 L

[ OH

−

[A [A

−

] =

][ HA ] −

=

Kw Ka

]

−3 7 . 789 ∞ 10 mol

0 . 0852 3 L

=

− 14 1 . ∞ 10

7 . 7 ∞ 10

−5

= 1 . 3 ∞ 10

− 10

= 0 . 0914 M

1988 D 12 X 10

X

8 pH

X X

6 X

X

X

4X 2 0 0

5

10 15 20 millilitres of NaOH

25

30

A 30.00 millilitre sample of a weak monoprotic acid was titrated with a standardized solution of NaOH. A pH meter was used to measure the pH after each increment of NaOH was added, and the curve above was constructed.

(a) Explain how this curve could be used to determine the molarity of the acid. (b) Explain how this curve could be used to determine the dissociation constant Ka of the weak monoprotic acid.

(c) If you were to repeat the titration using a indicator in the acid to signal the endpoint, which of the following indicators should you select? Give the reason for your choice. Ka = 1x10-5 Ka = 1x10-8 Ka = 1x10-11

Methyl red Cresol red Alizarin yellow

(d) Sketch the titration curve that would result if the weak monoprotic acid were replaced by a strong monoprotic acid, such as HCl of the same molarity. Identify differences between this titration curve and the curve shown above. Answer: (a) The sharp vertical rise in pH on the pH-volume curve appears at the equivalence point (about 23 mL). Because the acid is monoprotic, the number of moles of acid equals the number of moles of NaOH. That number is the product of the exact volume and the molarity of the NaOH. The molarity of the acid is the number of moles of the acid divided by 0.30L, the volume of the acid. (b) At the half-equivalence point (where the volume of the base added is exactly half its volume at the equivalence point), the concentration [HX] of the weak acid equals the concentration [X-] of its anion. Thus, in the equilibrium expression [H+] [X-]/[HX] = Ka, [H+] = Ka. Therefore, pH at the half-equivalence point equals pKa. (c) Cresol red is the best indicator because its pK a (about 8) appears midway in the steep equivalence region. This insures that at the equivalence point the maximum color change for the minimal change in the volume of NaOH added is observed. (d)

1989 A In an experiment to determine the molecular weight and the ionization constant for ascorbic acid (vitamin C), a student dissolved 1.3717 grams of the acid in water to make 50.00 millilitres of solution. The entire solution was titrated with a 0.2211 molar NaOH solution. The pH was monitored throughout the titration. The equivalence point was reached when 35.23 millilitres of the base has been added. Under the conditions of this experiment, ascorbic acid acts as a monoprotic acid that can be represented as HA.

(a) From the information above, calculate the molecular weight of ascorbic acid. (b) When 20.00 millilitres of NaOH had been added during the titration, the pH of the solution was 4.23. Calculate the acid ionization constant for ascorbic acid. (c) Calculate the equilibrium constant for the reaction of the ascorbate ion, A-, with water. (d) Calculate the pH of the solution at the equivalence point of the titration. Answer: (a) (0.2211M)(0.03523L) = 7.789x10-3 mol 1.3717g/7.789x10-3 mol = 176.1g/mol (b) at pH 4.23, [H+] = 8.0x10-8M [A

−

]=

−1 ( 0 . 02000 L )( 0 . 2211 mol _ L )

[ HA ] =

−1 ( 0 . 01523 L )( 0 . 2211 mol _ L )

X

Longer equivalence region for strong acid

10

X

= 0 . 04810 M

0 . 07000 L K =

+

[H

][ A

−

]

=

−5 ( 5 . 9 ∞ 10 ) ( 0 . 06317 )

[ HA ]

= 7 . 7 ∞ 10

−5

( 0 . 04810 )

(c) A + H2O <=> HA + OH-

K =

[ OH

−

[A 12

= 0 . 06317 M

0 . 07000 L

][ HA ] −

]

=

Kw Ka

=

1 . ∞ 10

− 14

7 . 7 ∞ 10

−5

= 1 . 3 ∞ 10

− 10

(d) at equiv. pt. [A

−

] =

−3 7 . 789 ∞ 10 mol

= 0 . 0914 M

0 . 08523 L 8 pH

X X

6 X X volume of NaOH 4 Same X added for equivalence point

X

2 0 0

5

Acid portion at lower pH for strong acid 10 15 20 25 30 millilitres of NaOH

[OH-]2 = (1.3x10-10)(9.14x10-2) = 1.2x10-11 [OH-] = 3.4x10-6M pOH =-log(3.4x10-6) =5.47; pH = (14-5.47)= 8.53 1994 D A chemical reaction occurs when 100. milliliters of 0.200-molar HCl is added dropwise to 100. milliliters of 0.100-molar Na 3P04 solution.

(a) Write the two net ionic equations for the formation of the major products.

(b) Identify the species that acts as both a Bronsted • Analytical balance acid and as a Bronsted base in the equation in • Phenolphthalein indicator solution (a), Draw the Lewis electron-dot diagram for this • Potassium hydrogen phthalate, KHP, a pure solid species. monoprotic acid (to be used as the primary (c) Sketch a graph using the axes provided, showing standard) the shape of the titration curve that results when (a) Briefly describe the steps you would take, using 100. milliliters of the HCl solution is added slowly the materials listed above, to standardize the from a buret to the Na3PO4 solution. for NaOH solution. the shape of the curve. (b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH solution. (c) After the NaOH solution has been standardized, it is used to titrate a weak monoprotic acid, HX. The equivalence point is reached when 25.0 mL of pH NaOH solution has been added. In the space provided at the right, sketch the titration curve, showing the pH changes that occur as the volume 0 of NaOH solution added increases from 0 to 35.0 mL HCl mL. Clearly label the equivalence point on the (d) Write the equation for the reaction that occurs if a curve. few additional milliliters of the HCl solution are added to the solution resulting from the titration in (c). Answer: (a) PO43- + H+ --> HPO42-; HPO42- + H+ --> H2PO4(b) HPO42. .– :O : . . . . . . O :H . . : : .P. : O . . :O . .: –

3–

+

2–

P O 4+ H ∅ H P O 4

(d) Describe how the value of the acid-dissociation constant, Ka, for the weak acid HX could be determined from the titration curve in part (c). (e) The graph below shows the results obtained by 0 mL HCl titrating a different weak acid, H2Y, with the (c) standardized NaOH solution. Identify the negative (d) H+ + H2PO4- --> H3PO4 ion that is present in the highest concentration at 1998 D (Required) [repeated in lab procedures section] the point in the titration represented by the letter A An approximately 0.1-molar solution of NaOH is to be on the curve. standardized by titration. Assume that the following materials are available. • Clean, dry 50 mL buret • 250 mL Erlenmeyer flask • Wash bottle filled with distilled water pH

2– + H PO 4 + H ∅

– H 2P O 4

In aqueous solution, ammonia reacts as represented above. In 0.0180 M NH3(aq) at 25ºC, the hydroxide ion concentration, [OH–] is 5.60x10–4 M. In answering the following, assume that temperature is constant at 25ºC and that volumes are additive. (a) Write the equilibrium-constant expression for the reaction represented above. (b) Determine the pH of 0.0180 M NH3(aq). Answer (a) • exactly mass a sample of KHP in the Erlenmeyer (c) Determine the value of the base ionization constant, Kb, of NH3(aq). flask and add distilled water to dissolve the solid. • add a few drops of phenolphthalein to the flask. (d) Determine the percent ionization of NH3 in • rinse the buret with the NaOH solution and fill. 0.0180 M NH3(aq). • record starting volume of base in buret. (e) In an experiment, a 20.0 mL sample of 0.0180 M • with mixing, titrate the KHP with the NaOH NH3(aq) was placed in a flask and titrated to the solution until it just turns slightly pink. equivalence point and beyond using 0.0120 M • record end volume of buret. HCl(aq). • repeat to check your results. (i) Determine the volume of 0.0120 M HCl(aq) mass of KHP that was added to reach the equivalence (b) molar mass KHP = moles of KHP point. since KHP is monoprotic, this is the number of (ii) Determine the pH of the solution in the flask moles of NaOH after a total of 15.0 mL of 0.0120 M HCl(aq) moles of NaOH L of titrant

was added. = molarity of NaOH

equivalence point

(iii) Determine the pH of the solution in the flask after a total of 40.0 mL of 0.0120 M HCl(aq) was added. Answer (a) Kb = (b) pOH = -log(5.60x10–4) = 3.252 pH = 14 – pOH = 10.748 (c) Kb = = 1.80x10–5 (d) x 100% = 3.11% (e) (i) NAVA = NBVB (0.0120 N)(VA) = (0.0180 N)(20.0 mL)

(c) (d) from the titration curve, at the 12.5 mL volume point, the acid is half-neutralized and the pH = pKa. Ka = 10pKa (e) Y2- (could it be OH- ?) 1999 A Required NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)

VA = 30.0 mL (ii) at 15.0 mL it is half-titrated and [NH 4+] = [NH3], then the Kb = [OH–] = 1.80x10–5. pOH = -log(1.80x10–5) = 4.745 pH = 14 – 4.75 = 9.255 (iii) at 40.0 mL, there is an excess of 10.0 mL of HCl past equivalence point,

(10.0 mL)(0.0120 M) = 0.120 mmol H+ ions = 0.00200M =[H+] pH = -log(0.00200) = 2.70