Question

question 15 (5 points)
how many moles of naoh is needed to neutralize 45.0 ml of 0.30m h₂seo₄?
a) 0.0135
b) 27.0
c) 0.00675
d) 0.027

Explanation:

Step1: Write the neutralization reaction

The reaction between \(H_2SeO_4\) (a diprotic acid) and \(NaOH\) (a strong base) is: \(H_2SeO_4 + 2NaOH
ightarrow Na_2SeO_4 + 2H_2O\). From the reaction, 1 mole of \(H_2SeO_4\) reacts with 2 moles of \(NaOH\).

Step2: Calculate moles of \(H_2SeO_4\)

Molarity \(M=\frac{n}{V}\) (where \(n\) is moles, \(V\) is volume in liters). Given \(V = 45.0\space mL=0.045\space L\), \(M = 0.30\space M\). So \(n_{H_2SeO_4}=M\times V = 0.30\space mol/L\times0.045\space L = 0.0135\space mol\).

Step3: Relate moles of \(H_2SeO_4\) to \(NaOH\)

From the reaction, the mole ratio of \(NaOH\) to \(H_2SeO_4\) is 2:1. So \(n_{NaOH}=2\times n_{H_2SeO_4}=2\times0.0135\space mol = 0.027\space mol\).

Answer:

D) 0.027