A standard solution is one that is produced so that the concentration is known. A standard solution is especially useful for the determination of the concentration of another solution with which it reacts. The procedure used is titration, in which the standard solution is added to a definite volume of the unknown solution until the chemical reaction is complete. The completion of the reaction can be detected by a product, such as a precipitate that forms, or by an indicator. Indicators are substances that are sensitive to a change in the solution and may signal completion of the reaction by changing color.
Two cautions about titrations are called for at this point. The first is that the volume of the unknown and the volume of the standard solution must be carefully measured so that the calculations will reflect what is actually present. The second caution deals with running a titration—make certain to perform titrations slowly so that the end-point of the titration is not exceeded. Going past the end-point may occur because the reaction is slower than the addition of drops or the end-point was simply missed because the drops are released faster than the color change can be perceived.
The goal of the titration procedure is to use up all of the unknown while keeping track of the amount of standard solution required for the titration. Once completed, the amount of the standard solution can be used in calculations to determine the concentration of the unknown solution. Determining the unknown by titration is based on the knowledge of the standard solution’s composition by
or
Sometimes, the calculation involves a monoprotic acid and a dihydroxy base or another set of conditions in which the relationship is not 1:1. We have to keep track of the various concentrations so that the molarities do not get mixed up. However, stoichiometric calculations involving solutions of specified normalities are even simpler. By the definition of equivalent mass in Chapter 12, two solutions will react exactly with each other if they contain the same number of equivalents. The calculation involved is
Suppose we have a standard solution and we need to determine information about a substance that will not dissolve in water. Solutions of given normalities are useful even when only one of the reactants is dissolved. In this case, the number of equivalents (or meq) of the reactant that is not in solution (not dissolved) is determined in the usual way, which is by dividing the mass of the sample in grams (or mg) by the equivalent mass. The reason that this works is that the number of equivalents (or meq) of one reactant must still equal the number of eq (or meq) of the other by the application of V1N1 = V2N2.
13.1. What volume of 1.40 M H2SO4 solution is needed to react exactly with 100 g Al?
As with any problem that includes the word react, the chance of needing a balanced chemical equation is pretty much a foregone conclusion. The single replacement reaction, written and balanced, is
Mole Method
Factor-Label Method
13.2. In standardizing a solution of AgNO3 it was found that 40.0 mL were required to precipitate all the chloride ions contained in 36.0 mL of 0.520 M NaCl. How many grams of silver could be obtained from 100 mL of AgNO3 solution?
The balanced equation tells us that equal numbers of moles AgNO3 and NaCl must be used. (As in previous chapters, n is the number of moles.)
Then, 40.0 mL of AgNO3 solution contains 0.01872 mol AgNO3, which provides 0.01872 mol Ag. Using these data, 100 mL of solution contains
13.3. Exactly 40.0 mL of 0.225 M AgNO3 were required to react with 25.0 mL of a solution of NaCN. Calculate the molarity of NaCN, if the reaction is by the following equation:
n(AgNO3) = (0.0400 L)(0.225 mol/L) = 0.00900 molAgNO3
and
n(NaCN) = 2 × n(AgNO3) = 0.0180 molNaCN
Then, 25.0 mL of the NaCN solution contains 0.0180 mol NaCN, so that
13.4. What volume in mL of 6.0 N NaOH is required to neutralize 30 mL of 4.0 N HCl?
Since HCl + NaOH → NaCl + H2O, and we have access to the formula, V1N1 = V2N2, we can simply substitute and solve for the desired normality.
then,
13.5. What is the normality of a H3PO4 solution if 40 mL neutralizes 120 mL of 0.531 N NaOH?
Since we are working with normalities, the solutes will react exactly with each other. Therefore,
Note: This problem’s solution works because we do not have to know whether one, two, or three hydrogens from H3PO4 are replaceable (or even the formula of the acid). The normality was determined by reaction of the acid with a base of known concentration. Under the concept of normality, the acid will have the same concentration (N) as the base, 1.59 N, in reactions with any strong base under similar conditions. In order to know the molar concentration of the acid, however, it would be necessary to know the number of replaceable hydrogens in the reaction, which we don’t.
In a case like this problem, where a substance can have several equivalent masses, the normality determined by one type of reaction is not necessarily the normality in other reactions. For instance, if a weak base like NH3 were to be used instead of a strong base for neutralizing the acid, or if the method of detecting the point of neutralization were changed (a different indicator), the equivalent mass of phosphoric acid (and the normality) might well be different.
13.6. (a) What volume of 5.00 N H2SO4 will neutralize a solution containing 2.50 g NaOH? (b) What mass in grams of pure H2SO4 is required for the reaction?
(a) One equivalent of H2SO4 reacts completely with one equivalent of NaOH. The equivalent mass of NaOH is the molar mass, 40.0. Then,
(b) The mass of acid is then calculated from the result of first calculation above.
13.7. A 0.250-g sample of solid acid was dissolved in water and neutralized by 40.0 mL of 0.125 N base. What is the equivalent mass of the acid?
13.8. Exactly 48.4 mL of HCl solution are required to neutralize completely 1.240 g pure CaCO3. Calculate the normality of the acid.
Each carbonate ion requires two hydrogen ions for neutralization by the reaction
Because of the relationship above, the equivalent mass of CaCO3 is 50.05, one-half the molar mass.
So, 48.4 mL of acid solution contains 0.0248 eq HCl
13.9. When 50 mL of a certain Na2CO3 solution was titrated with 0.102 M HCl, 56.3 mL were required for complete neutralization by the same reaction in Problem 13.8 in which two hydrogen ions are required per carbonate ion. Calculate the mass in grams of CaCO3 that would be precipitated if an excess of CaCl2 were added to a separate 50.0-mL portion of the Na2CO3 solution.
Factor-Label Method
13.10. A 10.0-g sample of “gas liquor” is boiled with an excess of NaOH and the resulting ammonia is passed into 60 mL of 0.90 N H2SO4. Exactly 10.0 mL of 0.40 N NaOH are required to neutralize the excess sulfuric acid (not neutralized by the NH3). What is the percent NH3 in the “gas liquor” examined?
In neutralization experiments, the equivalent mass of NH3 is the same as the molar mass, 17.0, conforming with the reaction 
Then, the mass of ammonia in the sample is (50 meq)(17.0 mg/meq) = 850 mg, which is 0.85 g ammonia. The fraction can be calculated by
13.11. A 40.8-mL sample of an acid is equivalent to 50.0 mL of a Na2CO3 solution, 25.0 mL of which is equivalent to 23.8 mL of 0.102 N HCl. What is the normality of the first acid?
The volume of HCl that would have been required for 50.0 mL of Na2CO3 solution is
13.12. Calculate the mass (g) of FeSO4 that will be oxidized by 24.0 mL of 0.250 N KMnO4 in a solution acidified with sulfuric acid. The unbalanced equation for the reaction is below. The statement of normality of KMnO4 is with respect to this reaction (Mn changes from +7 to +2 during this reaction).
It is not necessary to balance the equation. All we need to know is that the oxidation number on the iron changes from +2 to +3, as indicated by the charges on the iron ions. The equivalent mass FeSO4 is
Note: The same result would have been achieved if the balanced half-reaction had been used. That reaction is Fe2+ → Fe3+ + 1e–.
Let w = required mass of FeSO4.
13.13. What volume of 0.1000 N FeSO4 is required to reduce 4.000 g KMnO4 in a solution acidified with sulfuric acid?
The normality of the FeSO4 is with respect to the oxidation-reduction reaction in Problem 13.12. In this reaction, the Mn changes in oxidation number from +7 to +2. The net change is 5. Or, from the balanced half-reaction,
it can be seen that the electron transfer is 5 for each permanganate ion. The equivalent mass of KMnO4 in this reaction is
13.14. A sample known to contain As2O3 was brought into solution by a process which converted the arsenic to H3AsO3. This was titrated with a standard I2 solution according to the equation
Exactly 40.27 mL of standard I2 solution were required to reach the end-point as indicated by the persistence of a faint I2 color. The standard solution had been prepared by mixing 0.4192 g of pure KIO3 with excess KI and acid, then dissolving and diluting to 250.0 mL. The I2 is formed during the reaction,
Calculate the mass of As2O3 in the sample.
First, it is necessary to calculate the molar concentration of the I2 solution. By the factor-label method, the calculation is
Then, by the same method,
The above setup provides us with 0.09363 g As2O3.
13.15. What is the volume in mL of 0.25 M AgNO3 required to precipitate all the chromate ion from 20 mL of a solution containing 100 g of Na2CrO4 per liter? The reaction is
Ans. 99 mL
13.16. A 25-mL sample of hydrochloric acid that is to be used in treated swimming pools requires 44.2 mL of 6 M NaOH for complete neutralization. (a) What is the molarity of the HCl solution? (b) Calculate the weight/weight percentage of HCl; the mass of the sample is 25.00 g.
Ans. (a) 10.6 M; (b) 38.6%
13.17. A 50.0-mL sample of Na2SO4 solution is treated with an excess of BaCl2. If the precipitated BaSO4 is 1.756 g, what is the molarity of the Na2SO4 solution?
Ans. 0.1505 M
13.18. What was the thorium content of a sample that required 35.0 mL of 0.0200 M H2C2O4 to precipitate all the thorium as Th(C2O4)2?
Ans. 81 mg
13.19. Ba2+ is suspected to be in some effluent water downstream from an industrial installation. A 25-mL sample of the water is titrated with 0.35 M Na2SO4 solution until the insoluble BaSO4 is no longer produced, which required 53 mL Na2SO4 solution. Provide the concentration of barium in the water in terms of (a) molarity and (b) g Ba2+/mL
Ans. (a) 0.742 M; (b) 0.10 g/mL
13.20. What is the molar concentration of a K4Fe(CN)6 solution if 40.0 mL were required to titrate 150.0 mg Zn (dissolved) by forming K2Zn3[Fe(CN)6]2?
Ans. 0.0382M
13.21. A 50.0-mL sample of NaOH solution requires 27.8 mL of 0.100 N acid in a titration. (a) What is its normality? (b) How many mg NaOH are in each cubic centimeter (mL)?
Ans. (a) 0.0556 N; (b) 2.22 mg/cm3
13.22. In standardizing HCl, 22.5 mL were required to neutralize 25.0 mL of 0.0500 M Na2CO3 solution. What is the molarity and the normality of the HCl solution? How much water must be added to 200 mL of it to make it 0.100 N?
Ans. 0.111 M; 0.111 N; 22 mL
13.23. Exactly 21 mL of 0.80 N acid were required to neutralize completely 1.12 g of an impure sample of calcium oxide. What is the purity of the CaO?
Ans. 42%
13.24. Sulfuric acid is one of the most important acids used in industry and knowledge of its purity is extremely important. Find the (a) normality of a 35-mL sample of H2SO4 solution that requires 46 mL of 0.5000 M NaOH solution for complete neutralization; also, (b) state the concentration in terms of molarity.
Ans. (a) 0.66 N; (b) 0.33 M
13.25. Mg(OH)2 must have a purity of over 96% to be usable for a certain process. A 5.000 g sample requires 60.60 mL of 0.9000 M HCl for neutralization. The impurity has been found to be MgCl2; calculate the percent composition Mg(OH)2 by mass.
Ans. 31.8% Mg(OH)2
13.26. By the Kjeldahl method, the nitrogen contained in a foodstuff is converted into ammonia. If the ammonia from 5.0 g of a foodstuff is just sufficient to neutralize 20 mL of 0.100 M nitric acid, calculate the percentage of nitrogen in the foodstuff.
Ans. 0.56%
13.27. What is the purity of concentrated H2SO4 (density 1.800 g/cm3) if 5.00 cm3, after diluting in water, is neutralized by 84.6 mL of 2.000 M NaOH?
Ans. 92.2%
13.28. A 10.0-mL portion of (NH4)2SO4 solution was treated with an excess of NaOH. The NH3 gas evolved was absorbed in 50.0 mL of 0.100 M HCl. To neutralize the remaining HCl, 21.5 mL of 0.098 M NaOH were required. (a) What is the molar concentration of the (NH4)2SO4? (b) How many grams of (NH4)2SO4 are in a liter of solution?
Ans. (a) 0.145 M; (b) 19.1 g/L
13.29. Exactly 400 mL of an acid solution, when acted upon by an excess of zinc, evolved 2.430 L of H2 gas measured over water at 21°C and 747.5 torr. What is the normality of the acid? Vapor pressure of water at 21°C is 18.6 torr.
Ans. 0.483 N
13.30. How many grams of Cu will be replaced from 2.0 L of 0.150 M CuSO4 solution by 2.7 g of Al?
Ans. 9.5 g
13.31. What volume of 1.50 M H2SO4 will liberate 185 L of H2 gas at S.T.P. when treated with an excess of zinc?
Ans. 5.51 L
13.32. How many liters of H2 at S.T.P. would be replaced from 500 mL of 3.78 M HCl by 125 g Zn?
Ans. 21.2 L
13.33. The mass of 1.243 g of an acid is required to neutralize 31.72 mL of 0.1923 N standard base. What is the equivalent mass of this acid?
Ans. 203.8 g/eq
13.34. The molar mass of an organic acid was determined by the following study of its barium salt. 4.290 g of the salt were to be converted to the free acid by reaction with 21.64 mL of 0.477 M H2SO4. The barium salt was known to contain 2 moles water of hydration per mol Ba2+, and the acid was known to be monoprotic. What is the molar mass of the anhydrous acid?
Ans. 122.1 g/mol
13.35. An FeSO4 solution was standardized by titration. A 25.00-mL portion of the solution required 42.08 mL of 0.0800 N ceric sulfate for complete oxidation. What is the normality of the iron(II) sulfate?
Ans. 0.1347 N
13.36. A 15-g sample of tissue is suspected of containing arsenic. It is treated to convert any arsenic present to As(NO3)3 and diluted to 500 mL. A titration is performed using 0.0050 M H2S solution to precipitate the arsenic as As2S3. The volume of 0.53 mL (a little less than 11 drops at 20/mL) is required. How much arsenic (ppm) was present in the tissue?
Ans. 8.9 ppm arsenic
13.37. How many mL of 0.0257 N KIO3 would be needed to reach the end-point in the oxidation of 34.2 mL of 0.0416 N hydrazine in hydrochloric acid solution?
Ans. 55.4 mL
13.38. How many grams of FeCl2 will be oxidized by 28 mL of 0.25 N K2Cr2O7 in HCl solution? The unbalanced equation is
Ans. 0.89 g
13.39. In a standardization procedure, 13.76 mL of iron(II) sulfate solution were required to reduce 25.00 mL of potassium dichromate solution, which was prepared by dissolving 1.692 g of K2Cr2O7 in water and diluting to 500.0 mL. (See reaction in Problem 13.38.) Calculate the molarity and the normality of both the potassium dichromate and the iron(II) sulfate solution.
Ans. K2Cr2O7 0.01150 M; 0.06901 N; FeSO4 0.1254 M; 0.1254 N
13.40. What mass of MnO2 is reduced by 35 mL of 0.080 M oxalic acid, H2C2O4, in sulfuric acid solution? The unbalanced equation is
Ans. 0.24 g
13.41. (a) What mass of KMnO4 is required to oxidize 2.40 g FeSO4 in a solution acidified with sulfuric acid? (a) What is the equivalent mass of KMnO4 in this reaction?
Ans. (a) 0.500 g; (b) 31.6 g/eq
13.42. Find the equivalent mass of KMnO4 in the reaction
How many grams of MnSO4 are oxidized by 1.25 g KMnO4?
Ans. 52.7 g/eq, 1.79 g
13.43. (a) What volume of 0.0667 M K2Cr2O7 is required to liberate the chlorine from 1.20 g of NaCl in a solution acidified with H2SO4?
(b) How many grams of K2Cr2O7 are required? (c) How many grams of chlorine are liberated?
Ans. (a) 51 mL; (b) 1.01 g; (c) 0.73 g
13.44. If 25.0 mL of an iodine solution is equivalent as an oxidizing agent to 0.125 g of K2Cr2O7, to what volume should 1000 mL be diluted to make a solution 0.0500 M? The half-reaction for the iodine is
I2 + 2e → 2I–
Ans. 1020 mL
13.45. How many grams of KMnO4 should be taken to make up 250 mL of a solution of such concentration that 1 mL is equivalent to 5.00 mg iron in FeSO4?
Ans. 0.707 g
13.46. What mass of iodine is in a solution which requires 40 mL of 0.112 M Na2S2O3 to react with it?
Ans. 0.57 g
13.47. All the manganese in a certain solution was converted to 
ion by contact with solid sodium bismuthate. A 25.00-mL portion of 0.0200 M FeSO4 was added, which was more than sufficient to completely reduce the permanganate to Mn2+ in an acidified medium. The excess Fe2+ was then back-titrated in an acid solution, requiring 4.21 mL of 0.0106 M KMnO4. How many milligrams of manganese were in the original solution?
Ans. 3.04 mg
13.48. Reducing sugars are sometimes characterized by a number, RCu, which is defined as the number of milligrams of copper reduced by 1 gram of the sugar, in which the half-reaction for the copper is
It is sometimes convenient to determine the reducing power of a carbohydrate by an indirect method. In this method, 43.2 mg of the carbohydrate were oxidized by an excess of K3Fe(CN)6. The 
formed in this reaction required 5.29 mL of 0.0345 N Ce(SO4)2 for reoxidation to 
(the normality of the ceric sulfate solution is given with respect to the reduction of Ce4+ to Ce3+). Determine the RCu-value for the sample. (Hint: The number of meq Cu in a direct oxidation is the same as the number of meq Ce4+ in the indirect method.)
Ans. 268
13.49. A volume of 12.53 mL of 0.05093 M selenium dioxide, SeO2, reacted exactly with 25.52 mL of 0.1000 M CrSO4. In the reaction, Cr2+ was oxidized to Cr3+. To what oxidation number was the selenium converted by the reaction?
Ans. 0
13.50. A 150 g sample of ore is treated with nitric acid to remove the copper in the form of CuCl2. The resultant 587 mL of solution were titrated with 0.75 M NaOH to precipitate Cu(OH)2, which requires 41.7 mL. What is the yield from the ore expressed in terms of kg Cu/metric ton ore (1000 kg)?
Ans. 6620 kg Cu/metric ton ore
13.51. An acid solution of KReO4 sample containing 26.83 mg of combined rhenium was reduced by passage through a column of granulated zinc. The effluent solution, including the washings from the column, was then titrated with 0.1000 N KMnO4. A volume of 11.45 mL of the standard permanganate was required for the reoxidation of all the rhenium to the perrhenate ion, 
Assuming that rhenium was the only element reduced, what was the oxidation number to which rhenium was reduced by the zinc column?
Ans. –1
13.52. The iodide content of a solution was determined by titration with ceric sulfate in the presence of HCl, in which I– was converted to ICl and Ce4+ to Ce3+. A 250-mL sample of the solution required 20.0 mL of 0.050 M Ce4+ solution. What is the iodide concentration in the original solution (mg/mL)?
Ans. 0.25 mg/mL
13.53. An alloy of silver and copper can be used to make coins. A 75.00-g sample is treated with sufficient HCl to just precipitate all of the silver present as AgCl. The acid required is 16.32 mL of 6.25 M HCl. What is the percentage copper in the alloy?
Ans. 86.4%
13.54. A 0.518-g sample of limestone is dissolved, and then the calcium is precipitated as calcium oxalate, CaC2O4. The precipitate is filtered, washed, and dissolved in sulfuric acid solution. It then requires 40.0 mL of 0.0500 M KMnO4 solution to titrate it. What is the percent CaO in the limestone? The unbalanced equation for the titration is
Ans. 54.2%
13.55. Cadaverine is an organic compound that has a strong, disagreeable odor. Treatment of 15.00 mL of a smelly solution with 1.75 mL of 5 M HCl wiped out the odor (your nose was the indicator). (a) What was the normality of the smelly solution? (b) Why didn’t this question ask for molarity, rather than normality?
Ans. (a) 0.58N. (b) Molarity of the solution depends on the number of H+ ions that can be accepted by each molecule of cadaverine and there is no hint given to provide that information.
13.56. Assume the reaction between phosphate ions and calcium ions goes to completion to form Ca3(PO4)2. If 50.0 mL of 0.400 M Ca(NO3)2 are mixed with 100.0 mL of 0.300 M Na3PO4, what will be the final molarity of Ca2+, 
, Na+, and 
in the resulting solution?
Ans. no Ca2+, 0.267 M , 0.600 M Na+, 0.111 M
13.57. How many milliliters of 0.0876 M NaOH should be added to 50.0 mL of 0.0916 M HCl to yield a solution in which the H+ concentration is 0.0010 M?
Ans. 51.1 mL
13.58. One problem with standard NaOH solutions is that they absorb CO2 from the air, which reacts with OH– to form 
obscuring the end-point in titrations of acid. An enterprising young chemist had 975.0 mL of 0.3664 M NaOH, which she suspected had absorbed some CO2, so she added 10 mL of 0.500 M Ba(OH)2. That addition precipitated all the carbonate ion as BaCO3(s). After the BaCO3 was filtered off, the solution was restandardized and found to be 0.3689 M OH–. How many grams of solid BaCO3 had been removed?
Ans. 0.38 g
13.59. Citric acid, like other acids, tastes sour and can cause orange juice to be sour. Assuming that citric acid is the only acid present, a liter-size sample is titrated with 193 mL of 0.75 M NaOH. Also, assuming that citric acid is the only component in the juice reacting, what is the normality of citric acid present?
Ans. 0.15 N
13.60. A solution suspected of containing lead(II) nitrate is titrated with 0.035 M Na2SO4. How much Na2SO4 solution is required to produce a precipitate of PbSO4 from 1 L of a saturated lead(II) sulfate solution? The solution is saturated at 1.3 × 10–4 M PbSO4?
Ans. Over 3.7 mL added will cause a precipitate to appear.