Question

(A) A 14.5 g sample of an aqueous solution of hydrobromic acid contains an unknown amount of the acid.
If 26.8 mL of 0.557 M barium hydroxide are required to neutralize the hydrobromic acid, what is the percent by mass of hydrobromic acid in the mixture?

(B) A 13.7 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid.
If 19.8 mL of 0.682 M potassium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture?

(C) A 8.62 g sample of an aqueous solution of perchloric acid contains an unknown amount of the acid.
If 15.5 mL of 4.56×10^-2 M barium hydroxide are required to neutralize the perchloric acid, what is the percent by mass of perchloric acid in the mixture?

Answer 1

A)

Balanced chemical equation is:

Ba(OH)2 + 2 HBr ---> BaBr2 + 2 H2O

lets calculate the mol of Ba(OH)2

volume , V = 26.8 mL

= 2.68*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.557*2.68*10^-2

= 1.493*10^-2 mol

According to balanced equation

mol of HBr reacted = (2/1)* moles of Ba(OH)2

= (2/1)*1.493*10^-2

= 2.986*10^-2 mol

This is number of moles of HBr

Molar mass of HBr,

MM = 1*MM(H) + 1*MM(Br)

= 1*1.008 + 1*79.9

= 80.908 g/mol

use:

mass of HBr,

m = number of mol * molar mass

= 2.986*10^-2 mol * 80.91 g/mol

= 2.416 g

mass % = mass of HBr * 100 / mass of solution

= 2.416*100/14.5

= 16.7 %

Answer: 16.7 %

B)

Balanced chemical equation is:

KOH + HNO3 ---> KNO3 + H2O

lets calculate the mol of KOH

volume , V = 13.7 mL

= 1.37*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.682*1.37*10^-2

= 9.343*10^-3 mol

According to balanced equation

mol of HNO3 reacted = (1/1)* moles of KOH

= (1/1)*9.343*10^-3

= 9.343*10^-3 mol

This is number of moles of HNO3

Molar mass of HNO3,

MM = 1*MM(H) + 1*MM(N) + 3*MM(O)

= 1*1.008 + 1*14.01 + 3*16.0

= 63.018 g/mol

use:

mass of HNO3,

m = number of mol * molar mass

= 9.343*10^-3 mol * 63.02 g/mol

= 0.5888 g

Use:

mass % = mass of HNO3 * 100 / mass of solution

= 0.5888*100/13.7

= 4.30 %

Answer: 4.30 %

C)

Balanced chemical equation is:

Ba(OH)2 + 2 HClO4 ---> Ba(ClO4)2 + 2 H2O

lets calculate the mol of Ba(OH)2

volume , V = 15.5 mL

= 1.55*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 4.56*10^-2*1.55*10^-2

= 7.068*10^-4 mol

According to balanced equation

mol of HClO4 reacted = (2/1)* moles of Ba(OH)2

= (2/1)*7.068*10^-4

= 1.414*10^-3 mol

This is number of moles of HClO4

Molar mass of HClO4,

MM = 1*MM(H) + 1*MM(Cl) + 4*MM(O)

= 1*1.008 + 1*35.45 + 4*16.0

= 100.458 g/mol

use:

mass of HClO4,

m = number of mol * molar mass

= 1.414*10^-3 mol * 1.005*10^2 g/mol

= 0.142 g

Mass % = mass of HClO4 * 100 / mass of solution

= 0.142 * 100 / 8.62

= 1.65 %

Answer: 1.65 %