Answer:
0.36 N
Explanation:
Let's consider the neutralization between HBr and NaOH.
HBr + NaOH → NaBr + H₂O
The equivalents of NaOH that reacted are:
24.0 × 10⁻³ L × 1.5 eq/L = 0.036 eq
Since the reactions are equivalent to equivalent, also 0.036 equivalents of HBr reacted.
The normality of HBr is:
N = 0.036 eq / 0.100 L = 0.36 N
