A. An aqueous Solution of barium hydroxide is standardized by titration with a .

Question

A. An aqueous Solution of barium hydroxide is standardized by titration with a .135 in solution of hydrochloric acid. If 17.2 mL of base are required to neutralize 16.3 mL of the acid what is the molarity of the barium hydroxide solution
B. An aqueous solution of hydrochloric acid is standardized by titration with a .119M solution of barium hydroxide. If 19.6 mL of base are required to neutralize 24.4 mL of the acid what is the molarity of the HCl solution C. An aqueous solution of calcium hydroxide is standardized by titration with a .174 M solution of hydroiodic acid. If 24.6 mL of base are required to neutralize 19.9 mL of the acid what is the molarity of the calcium hydroxide solution?
A. An aqueous Solution of barium hydroxide is standardized by titration with a .135 in solution of hydrochloric acid. If 17.2 mL of base are required to neutralize 16.3 mL of the acid what is the molarity of the barium hydroxide solution
B. An aqueous solution of hydrochloric acid is standardized by titration with a .119M solution of barium hydroxide. If 19.6 mL of base are required to neutralize 24.4 mL of the acid what is the molarity of the HCl solution C. An aqueous solution of calcium hydroxide is standardized by titration with a .174 M solution of hydroiodic acid. If 24.6 mL of base are required to neutralize 19.9 mL of the acid what is the molarity of the calcium hydroxide solution?

B. An aqueous solution of hydrochloric acid is standardized by titration with a .119M solution of barium hydroxide. If 19.6 mL of base are required to neutralize 24.4 mL of the acid what is the molarity of the HCl solution C. An aqueous solution of calcium hydroxide is standardized by titration with a .174 M solution of hydroiodic acid. If 24.6 mL of base are required to neutralize 19.9 mL of the acid what is the molarity of the calcium hydroxide solution?

Explanation / Answer

a) from neutralisation concept

Ba(OH)2(aq) + 2HCl(aq) -----> BaCl2(aq) ++ 2H2O(l)

1 mol Ba(OH)2(aq) = 2 mol HCl(aq)

M1V1/n1 = M2V2/n2

(M1*17.2/1) = (0.135*16.3/2)

M1 = molarity of Ba(OH)2(aq) = 0.064 M

b)

M1V1/n1 = M2V2/n2

(M1*24.4/2) = (0.119*19.6/1)

M1 = molarity of HCl(aq) = 0.1912 M

c)

Ca(OH)2(aq) + 2HCl(aq) -----> CaCl2(aq) ++ 2H2O(l)

1 mol Ca(OH)2(aq) = 2 mol HCl(aq)

M1V1/n1 = M2V2/n2

(M1*24.6/1) = (0.174*19.9/2)

M1 = molarity of Ca(OH)2(aq) = 0.07038 M

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