Question

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?

Answer 1

To find the percent by mass of hydrobromic acid in the mixture, we need to use the concept of titration. We can calculate the moles of hydrobromic acid using the volume and concentration of the barium hydroxide solution used to neutralize the acid. Then, we can determine the percent by mass.

Step-by-step explanation:

To find the percent by mass of hydrobromic acid in the mixture, we need to use the concept of titration. In a neutralization reaction, the moles of acid can be determined by multiplying the volume of the base solution with its concentration. This can be expressed using the formula:

Moles of acid = Volume of base solution (L) x Concentration of base solution (M)

In this case, we are given the volume and concentration of the barium hydroxide solution used to neutralize the hydrobromic acid. So, we can calculate the moles of hydrobromic acid and then determine the percent by mass.

Answer 2

3.3 %

Step-by-step 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 %