A 7.52 g sample of an aqueous solution of hydrobromic acid contains an unknown amount of the acid.
If 13.5 mL of 0.633 M barium hydroxide are required to neutralize the hydrobromic acid, what is the percent by mass of hydrobromic acid in the mixture?
To find the percent by mass of hydrobromic acid in the mixture, we need to calculate the amount of hydrobromic acid that was neutralized by the barium hydroxide.
First, we need to find the number of moles of barium hydroxide used. Given that the concentration of the barium hydroxide solution is 0.633 M and the volume used is 13.5 mL, we can calculate the moles of barium hydroxide:
Moles of barium hydroxide = concentration × volume
= 0.633 mol/L × 0.0135 L
≈ 0.00855 mol
From the balanced chemical equation between hydrobromic acid (HBr) and barium hydroxide (Ba(OH)2), we know that 1 mole of hydrobromic acid reacts with 2 moles of barium hydroxide. Therefore, the moles of hydrobromic acid neutralized can be calculated:
Moles of hydrobromic acid neutralized = 0.00855 mol/2
= 0.00427 mol
Now that we know the moles of hydrobromic acid that were neutralized, we can calculate the mass of hydrobromic acid using the molar mass of HBr.
The molar mass of HBr = atomic mass of hydrogen (1 g/mol) + atomic mass of bromine (79.9 g/mol)
= 1 g/mol + 79.9 g/mol
= 80.9 g/mol
Mass of hydrobromic acid = moles of hydrobromic acid × molar mass of HBr
= 0.00427 mol × 80.9 g/mol
≈ 0.345 g
Finally, we can calculate the percent by mass of hydrobromic acid in the mixture using the given mass of the solution (7.52 g):
Percent by mass of hydrobromic acid = (mass of hydrobromic acid / mass of the solution) × 100
= (0.345 g / 7.52 g) × 100
≈ 4.59%
Therefore, the percent by mass of hydrobromic acid in the mixture is approximately 4.59%.