A 25.0 mL solution of HCl is neutralized with 16.7 mL of 0.115 M Sr(OH)₂. What is the concentration of the original HCl solution?

Answer 1

The first thing to do is to write the correctly balanced equation:

2HCl + Sr(OH)₂ ==> SrCl₂ + 2H2O .. balanced equation

Next, we find the moles of Sr(OH)2 that were used to neutralize the HCl

16.7 ml x 1 L / 1000 ml x 0.115 mol / L = 0.00192 mols Sr(OH)2

Next, we use the mole ratios in the balanced equation to find mols of HCl present;

0.00192 mols Sr(OH)2 x 2 mols HCl / 1 mol Sr(OH)2 = 0.00384 mols HCl

Finally, we divide the mols of HCl by the volume to get mols/L or M, the original concentration:

0.00384 mols / 25.0 ml x 1000 mls / L = 0.154 M (3 sig. figs.)