A 14.1 g sample of an aqueous solution of hydrochloric acid contains an unknown

Question

A 14.1 g sample of an aqueous solution of hydrochloric acid contains an unknown amount of the acid. If 25.8 mL of 6.55×10-2 M barium hydroxide are required to neutralize the hydrochloric acid, what is the percent by mass of hydrochloric acid in the mixture?

A 13.2 g sample of an aqueous solution of perchloric acid contains an unknown amount of the acid. If 18.9 mL of 0.619 M barium hydroxide are required to neutralize the perchloric acid, what is the percent by mass of perchloric acid in the mixture?

Explanation / Answer

1)

m = 14.1 g

HCl

if

V = 25.8 ml and ; = 6.55*10^-2 M of Ba(OH)2 is used

then

Ba(OH)2 + 2HCl = 2H2O + BaCl2

ratio is 1 base : 2 acid so

mol base = M*V = 25.8*(6.55*10^-2) = 1.6899 mmol

then there are

2*1.6899 = 3.3798 mmol of acid

mass acid = mol*MW = (3.3798*10^-3)(36) = 0.1216728 g of acid

% mass = mass of acid / totall mass = 0.1216728 /14.1 * 100% = 0.8629 %

2)

m = 13.2 g of HClO4

V = 18.9 ml of M = 0.619 Ba(OH)"

ratio is also 1:2 so

mol of base = MV = 18.9*0.619 = 11.6991 mmol of base

then we have 11.6991*2 = 23.3982 mmol of acid

mass = mol*MW = (23.3982*10^-3)(100.46 )= 2.3505

%mass = 2.3505 / (13.2) * 100 = 17.8068 %

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