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

HBr is an acid (identified by the presence of H)

Sr(OH)₂ is a base (identified by the presence of OH)

When an acid and base are neutralized (by being mixed together), that laboratory process is called a titration.

The Titration Equation (provided on Table T of the Chemistry Reference Tables) can be used to solve this problem:

M_AV_A = M_BV_B

(concentration of acid)(volume of acid) = (concentration of base)(volume of base)

Here's the information we know for each variable:

M_A = unknown concentration of acid

V_A = 25.0 mL

M_B = 0.105 M

V_B = 22.6 mL

Now we plug in the information and solve for the unknown:

M_AV_A = M_BV_B

M_A(25.0) = (0.105)(22.6)

M_A = [(0.105)(22.6)] ÷ 25.0

M_A = 0.09492 ≈ 0.0949 M (answer rounded to 3 significant figures)

I hope this helps!