Question

A 7.53 g sample of an aqueous solution of hydrobromic acid contains an unknown amount of the acid.
If 19.8 mL of 0.483 M barium hydroxide are required to neutralize the hydrobromic acid, what is the percent by mass of hydrobromic acid in the mixture?

% by mass

Answer 1

SOLUTION

Moles of Barium hydroxide [(Ba(OH)₂] required = Molarity X Volume = 0.483 M X 19mL = 9.177 mmoles

Each mole of Ba(OH)₂ produces two moles of OH^- as

Ba(OH)₂ -------> Ba²⁺ + 2 OH^-

Hence moles of OH^- produced by 9.177 mmoles of Ba(OH)₂ = 9.177 X 2 = 18.354 mmoles.

Now one mole of OH^- is required to neutralise one mole of hydrobromic acid (HBr) as:

HBr + OH^- --------> Br- + H2O

Hence 18.354 mmoles of OH^- are required for 18.354 mmoles of HBr. In other words the 7.53g of aqueous solution contain 18.354 mmoles of HBr.

Mass of HBr present in 7.53g of aqueous solution = Molar mass of HBr X number of moles

= 81 X 18.354 mmol = 1486.6 mg = 1.5g

Percentage of HBr = (mass of HBr / Total mass of solution ) X 100 = (1.5 / 7.53) X 100 = 19.92%

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