Question

a. A 12.5 g sample of an aqueous solution of HBr contains an unknown amount of the acid.
If 30.7 mL of 0.850 M barium hydroxide are required to neutralize the HBr, what is the percent by mass of HBr in the mixture?

unbalanced equation: Ba(OH)₂ (aq) + HBr (aq) ----> H₂O (l) + BaBr₂ (aq)

b. write the net ionic equation (include phases and the charge of ions).

Answer 1

The balanced equation can be written as

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

i.e., 2 mol HBr is required to react with one mole Ba(OH)₂

Based on the given data, moles of Ba(OH)₂ = molarity * volume in L

Molarity = 0.850 M

Volume = 30.7 ml = 30.7 * 10^-3 L

moles of Ba(OH)₂ = 0.850 M * 30.7 * 10^-3 L = 0.026095 mol

Based on balanced eqution, 2 * 0.026095 mol of HBr is required.

So, moles of HBr = 2 * 0.026095 mol= 0.05219 mol

We know, molar mass of HBr =81 g/mol

So, mass of 0.05219 mol of HBr = 81* 0.05219 mol = 4.227 g

Now, percent by mass of HBr in the mixture = (4.227/12.5)* 100 = 33.816%

Ionic equation is given below: