Normality Problems

1. What is the normality of the following?

a. 0.1381 M NaOH

0.1381 mol/L x (1 eq/1mol) = 0.1381 eq/L = 0.1381 N

b. 0.0521 M H₃PO₄

0.0521 mol/L x (3 eq/1mol) = 0.156 eq/L = 0.156 N

c. 0.5781 g acid (eq wt = 187.3) in 250.0 mL of solution

0.5781 g x (1 eq/187.3 g) = 0.003086 eq/0.2500L = 0.01235 N

d. 0.321 g sodium carbonate in 250.0 mL of solution

0.321 g Na₂CO₃ x (1 mol/105.99 g) x (2 eq/1mol) = 0.01886 eq/0.2500L = 0.0755 N

2. What is the molarity of the following?

a. 0.3181 N HCl

0.3181 eq/L x (1 mol/1 eq) = 0.3181 M

b. 0.1115 N Sr(OH)₂

0.1115 eq/L x (1 mol/2 eq) = 0.05575 M

c. 0.4237 g of KHP (eq wt = 204.23) in 100.0 mL of solution

0.4237 g KHP x (1 eq/204.3 g) x (1 mol/1 eq) = 0.002074 mol/0.1000 L = 0.02074 M

3. What is the equivalent weight of the following?

a. C₆H₅COOH b. Na₂CO₃ c. H₃PO₄

a. mw = 122.13 g/mol x (1 mol/1 eq) = 122.13 g/eq

b. mw = 105.99 g/mol x (1 mol/2 eq) = 52.99 g/eq

c. mw = 98.00 g/mol x (1mol/3 eq) = 32.67 g/eq

4. If 25.00 mL of citric acid solution is titrated with 28.12 mL of 0.1718 N KOH, what is the concentration of citric acid? (hint: normality is a concentration unit!)

N_a·V_a = N_b·V_b N_a(25.00 mL) = (0.1718N)(28.12 mL) N_a = 0.1932 N

or

28.12 mL KOH x (0.1718 meq/1 mL) x (1 meq acid/1 meq base) = 4.831 meq acid/25.00 mL = 0.1932 N

5. If 31.87 mL of base is required in the standardization of 0.4258 g of KHP (eq wt = 204.23), what is the normality of the base?

0.4258 g KHP x (1 eq/204.23g) x (1 eq base/1eq acid) = 2.085 x 10^-3 eq base/0.03187 L = 0.6542 N

6. What is the normality of an acid if 21.18 mL were needed to titrate 0.1369 g Na₂CO₃?

0.1369 g Na₂CO₃ x (1 mol/105.99 g) x (2 eq/1 mol) x (1 eq acid/1 eq base) = 2.583 x 10^-3 eq acid/0.02118 L = 0.1212 N

7. What is the equivalent weight of a base is 0.2131 g required 43.71 mL of 0.0132 N acid?

43.71 mL x (1 L/1000 mL) x (0.0132 eq acid/1 L ) x (1 eq base/1 eq acid) = 5.770 x 10^-4 eq

eq wt = 0.2131 g/5.770 x 10^-4 eq = 369 g/eq

8. What is the %acid (eq wt = 173.8) if 20.07 mL of 0.1100 N base were required to neutralize 0.7231 g of sample?

20.07 mL x (1 L/1000 mL) x (0.1100 eq base/1 L) x (1 eq acid/1 eq base) x (173.8 g /1 eq) = 0.3837 g acid

(0.3837 g acid/0.7231 g sample) x 100 = 53.06%

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