Answer:
0.1738 M
Explanation:
Let's consider the neutralization reaction between NaOH and HBr.
NaOH + HBr → NaBr + H₂O
The moles of HBr are:
14.76 × 10⁻³ L × 0.4122 mol/L = 6.084 × 10⁻³ mol
The molar ratio of NaOH to HBr is 1:1. In the endpoint, they have reacted completely, so the moles of NaOH were 6.084 × 10⁻³ mol before the reaction.
The molarity of NaOH is:
M = 6.084 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.1738 M
