Question

# 1. An aqueous solution of barium hydroxide is standardized by titration with a 0.197 M solution...

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.

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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