1. An aqueous solution of barium hydroxide is standardized by titration with a 0
ID: 555084 • Letter: 1
Question
1. An aqueous solution of barium hydroxide is standardized by titration with a 0.197 M solution of hydrobromic acid. If 10.4 mL of base are required to neutralize 22.7 mL of the acid, what is the molarity of the barium hydroxide solution? M barium hydroxide
b.An aqueous solution of hydrochloric acid is standardized by titration with a 0.176 M solution of barium hydroxide.
If 13.7 mL of base are required to neutralize 14.1 mL of the acid, what is the molarity of the hydrochloric acid solution?
M hydrochloric acid
1c.
A 10.5 g sample of an aqueous solution of hydrobromic acid contains an unknown amount of the acid.
If 29.5 mL of 0.563 M potassium hydroxide are required to neutralize the hydrobromic acid, what is the percent by mass of hydrobromic acid in the mixture?
% by mass
d. A 13.5 g sample of an aqueous solution of hydrochloric acid contains an unknown amount of the acid.
If 25.1 mL of 0.643 M sodium hydroxide are required to neutralize the hydrochloric acid, what is the percent by mass of hydrochloric acid in the mixture?
% by mass
Identify the species oxidized, the species reduced, the oxidizing agent and the reducing agent in the following electron transfer reaction.
Explanation / Answer
1. Ba(OH)2 + 2HBr ---> BaBr2 + 2H2O
1 mol Ba(OH)2 = 2 mol HBr
a. M1V1/n1 = M2V2/n2
(M1*10.4/1) = (0.197*22.7/2)
M1 = 0.215 M
b. Ba(OH)2 + 2Hcl ---> Bacl2 + 2H2O
1 mol Ba(OH)2 = 2 mol HCl
M1V1/n1 = M2V2/n2
(0.176*13.7/1) = (M2*14.1/2)
M2 = 0.342 M
C. KOH + HBr ----> KBr + H2O
NO Of mol of KOH reacted = 0.563*29.5/1000 = 0.0166 mol
NO Of mol of HBr reacted = 0.0166 mol
mass of HBr reacted = 0.0166*81 = 1.345 g
%by mass of sample = 1.345/10.5*100 = 12.8%
d. KOH + HCl ----> KCl + H2O
NO Of mol of KOH reacted = 0.643*25.1/1000 = 0.0161 mol
NO Of mol of HCl reacted = 0.0161 mol
mass of HCl reacted = 0.0161*36.5 = 0.58765 g
%by mass of sample = 0.58765/13.5*100 = 4.35%
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