A 8.72 g sample of an aqueous solution of hydrobromic acid contains an unknown amount of the acid. If 19.5 mL of 0.374 M barium hydroxide are required to neutralize the hydrobromic acid, what is the percent by mass of hydrobromic acid in the mixture?

3.3 %

Explanation:

According to the question, the following reaction takes place -

Ba (OH) ₂ + 2 HBr → BaBr₂ + 2 H₂O

Molarity of a substance, is the number of moles present in a liter of solution.

M = n / V

M = molarity

V = volume of solution in liter,

n = moles of solute,

According to the question,

V = volume of Ba (OH) ₂ = 19.5 mL = 0.0195 L (since, 1 ml = 10 ⁻³ L)

M = Molarity of Ba (OH) ₂ = 0.374 M

The moles of Ba (OH) ₂ can be calculated by using the above equation,

M = n / V

n = M * V = 0.374 M * 0.0195 L = 0.0072 mol

From the above balanced reaction,

2 mol of HBr reacts with 1 mol Ba (OH) ₂

1 mol of HBr reacts with 1 / 2 mol Ba (OH) ₂

From the above data,

1 mol HBr reacts with = 1 / 2 * 0.0072 mol = 0.0036 mol

Hence, number of moles of HBr = 0.0036 mol

Now,

Moles is denoted by given mass divided by the molecular mass,

Hence,

n = w / m

n = moles,

w = given mass,

m = molecular mass.

As calculated above,

n = 0.0036 mol

As we know, the m = molecular mass of HBr = 81 g/mol

n = w / m

w = n * m = 0.0036 mol * 81 g/mol = 0.2916 g

Now,

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

mass % = 0.2916 g / 8.72 g * 100 = 3.3 %