Final answer:
To neutralize 60.0 mL of 0.0400 M HBr with 0.0200 M Sr(OH)₂, one needs to use 12000 mL of Sr(OH)₂, as determined by the stoichiometry of the balanced chemical equation and molarity-volume relationship.
Step-by-step explanation:
Calculating the Volume of Sr(OH)₂ to Neutralize HBr
To find the volume in milliliters of 0.0200 M Sr(OH)₂ required to neutralize 60.0 mL of 0.0400 M HBr, one needs to perform a titration calculation. The reaction between Sr(OH)₂ and HBr can be represented by the balanced chemical equation:
Sr(OH)₂ (aq) + 2HBr (aq) → SrBr₂ (aq) + 2H₂O (l)
The stoichiometry of the reaction indicates that one mole of Sr(OH)₂ reacts with two moles of HBr. Hence, we can write the molarity (M) and volume (V) relationship as:
(M1)(V1) = 2 * (M2)(V2)
Where:
M1 is the molarity of Sr(OH)₂ (0.0200 M).V1 is the volume of Sr(OH)₂ needed.M2 is the molarity of HBr (0.0400 M).V2 is the volume of HBr (60.0 mL or 0.0600 L).
Plugging the values into the equation:
(0.0200 M) * (V1) = 2 * (0.0400 M) * (0.0600 L)
Solving for V1 gives us:
V1 = (2 * 0.0400 M * 0.0600 L) / 0.0200 M
V1 = 0.2400 L / 0.0200 M
V1 = 12.00 L
However, since we are looking for the volume in milliliters, we convert L to mL:
V1 = 12.00 L * 1000 mL/L = 12000 mL
Therefore, 12000 mL of 0.0200 M Sr(OH)₂ is required to neutralize 60.0 mL of 0.0400 M HBr.