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Do the following conversions:
- If one completely neutralizes a 0.1500 M solution of Ca(OH) 2 , what is the initial normality of the Ca(OH) (^2) solution?
0.1500 M * 2 N = 0.3000 N 1 M ANS 0.3000 N
- H3PO4 is reacted with NaOH to form Na 3 PO 4. If one were to start with 3.50 equivalencies of H 3 PO4, how many mole of H 3 PO4 is this? 3.50 eq * 1 mol = 3 eq
ANS 1.17 mol
- A 0.1002 M solution of H 2 SO4 is reacted with NaOH to produce NaHSO 4 in a titration. If 24.5 mL of the H 2 SO4 solution is used, how many moles of H 2 SO 4 solution is used? H2SO4 + NaOH โ NaHSO 4 + H 2 O 0.1002 mol * 0.0245 L sol = 1 L sol
ANS 2.45x10-3^ mol
What is the molarity of a 1.035 N solution of HCl? 1.035 N * 1 M = 1 N ANS 1.035 M
For a monoprotic reaction, what is the normality of a 0.547 M solution of Ca(OH) 2? 1 N * 0.547 M 1 M ANS 0.547 N
How many grams of NaOH is contained in 0.500 mL of a 1.0 N solution? 40.0 g NaOH * 1 mol NaOH * 1.0 eq NaOH * 0.500x10-3^ L sol 1 mol NaOH 1 eq NaOH 1 L sol
ANS 2.0 x 10 -2^ g
- What is the normality of a solution of Mg(OH) 2 for a reaction in which both OH -^ are consumed if one has 1.34 mg of Mg(OH) 2 dissolved in 1.00 L of solution? 2 eq * 1 mol * 1 g * 1.34 mg * 1 1 mol 58.3 g 1000 mg 1.00 L sol
ANS 4.60x10 -5^ N
Calculate the following:
- For the reaction (unbalanced):
Cr2O72-^ + SO 3 2-^ โ Cr3+^ + SO42-
The balanced half reactions are:
14 H +^ + 6e -^ + Cr 2 O72-^ โ 2 Cr3+^ + 7H 2 O
H2O + SO 3 2-^ โ SO 4 2-^ + 2e -^ + 2H +
a. How many equivalencies of K 2 Cr2O7 is 38.22 g? MW K2Cr 2 O7 = 294.0 g/mol 6 eq * 1 mol * 38.22 g 1 mol 294.0 g
ANS 0.7794 eq
b. A 0.12 M solution of H 2 SO 3 is what normality? 2 N H 2 SO3 * 0.12 M H 2 SO 3 = 1 M H 2 SO
ANS 0.24 N
c. What is the molarity of a 0.1900 N K 2 Cr 2 O7 solution? 0.1900 N K 2 Cr 2 O 7 * 1 M K 2 Cr2 O7 = 6 N K 2 Cr 2 O
ANS 0.03167 M
d. If one has 0.200 moles of H 2 SO3, how many equivalencies does one have? 0.200 mol H 2 SO3 * 2 eq H 2 SO 3 = 1 mol H 2 SO
ANS 0.400 eq
Calculate the following:
- An unknown acid is titrated with a standardized base solution. The concentration of the base is 0.1022 N. 45.15 mL of the base solution was used to titrate 25.00 mL of the acid. What is the normality of the acid solution?
Using N (^) aVa = N (^) b Vb :
(25.00 mL) N (^) a = (0.1022 N )(45.15 mL) and solve for Na
ANS 0.1846 N
- In a titration to determine the concentration of Ca(OH) 2 in a solution, both OH -^ were neutralized. 10.00 mL of the Ca(OH) 2 solution were titrated with a standard acid. This standard acid was 0.1100 N and 23.98 mL of it was used for the titration. What is the molarity of the Ca(OH) (^2) solution?
Using N (^) aVa = N (^) b Vb :
(0.1100 N )(23.98 mL) = Nb(10.00 mL) Nb = 0.2638 N
1 M * 0.2638 N = 0.1319 M 2 N ANS 0.1319 M
- In the reaction between FeSO 4 (formula weight = 151.93 g/mol) and KMnO 4 , Fe2(SO 4 )3 and MnSO 4 are produced. A laboratory worker reacted 6.08 g of solid FeSO 4 with a solution of KMnO 4 until an endpoint is reached. 37.98 mL of the solution was used in the titration. What is the normality and molarity of the solution?
FW of FeSO 4 = 151.9 g/mol 1 eq FeSO4 = 1 eq KMnO
1 * 6.08 g FeSO 4 * 1 mol FeSO 4 * 1 eq FeSO 4 * 1 eq KMnO (^4) 0.03798 Lsol 151.9 g FeSO 4 1 mol FeSO 4 1 eq FeSO (^4)
ANS 1.054 N
1.054 N KMnSO 4 * 1 M KMnO 4 5 N KMnSO 4 ANS 0.2108 M
- An oxalic acid (H 2 C2 O4) solution was used to standardize a potassium permanganate solution in a redox titration. Before using the oxalic acid as a standard, however, it (oxalic acid) was standardized in a acid-base titration using NaOH solution. In the first (acid-base) titration, 25.00 mL of oxalic solution was used and 45.34 mL of NaOH solution of 0.1011 N was used. In the second (redox) titration, 25.00 mL of oxalic solution was used and 35.03 mL of permanganate solution was used.
a) Write the balanced equation for the acid-base reaction.
H 2 C2O 4 + 2NaOH โ Na 2 C 2 O4 + 2H 2 O
b) Calculate the normality and molarity of the oxalate in the acid-base case. Using N a V a = N b V b :
N a (25.00 mL) = (0.1011 N )(45.34 mL) and solving
ANS 0.1834 N diprotic therefore: 2 N = 1 M (unit factor or by formula N = 2 M ) ANS 0.09168 M
c) Write half reactions and the balanced equation for the redox reaction.
2H2O + C 2 O42-^ โ 2CO 3 2-^ + 2e -^ + 4H +
8H+^ + 5e -^ + MnO 4 -^ โ Mn2+^ + 4H 2 O
d) What is the normality of the oxalate in the redox case? 0.09168 M * 2 N = 0.1834 N 1 M
ANS 0.1834 N e) Calculate the normality and molarity of the permanganate in the redox case. Using N o V o = N r V r :
N o (35.03 mL) = (0.1834 N )(25.00 mL) N o = 0.1309 N
ANS 0.1309 N
0.1309 N * 1 M =
5 N
ANS 0.02618 M
