A 25.0 mL solution of HBr is neutralized with 22.6 mL of 0.105 M Sr(OH)₂. What is the concentration of the original HBr solution?

Answer 1

First, write the correctly balanced equation:

2HBr + Sr(OH)2 ==> SrBr2 + 2H2O

Next, find moles of Sr(OH)2 used:

22.6 ml x 1 L / 1000 ml x 0.105 mol / L = 0.002373 mols

Next, find moles of HBr this is equivalent to:

0.002373 mol Sr(OH)2 x 2 mol HBr / 1 mol Sr(OH)2 = 0.004746 mols

Finally, use this value to calculate the concentration of HBr:

0.004746 mol / 25.0 ml x 1000 ml / L = 0.190 mol / L (3 sig.figs.)