What volume of 0.5M H2SO4 will exactly neutralize 20cm3 of 0.1M NaOH Solution?
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Correct Answer: Option A
Explanation:
The balanced equation is given as:
\(H_{2}SO_{4} + 2NaOH \to Na_{2}SO_{4} + 2H_{2}O\)
Molarity of \(NaOH = 0.1M\)
Volume of \(NaOH = 20cm^{3}\)
Molarity of \(H_{2}SO_{4} = 0.5M\)
mol NaOH = \(0.1 \times 20cm^{3} = 2molNaOH\)
⇒ \(2mol NaOH \times \frac{1mol H_{2}SO_{4}}{2mol NaOH} = 1 mol H_{2}SO_{4}\)
\(volume of H_{2}SO_{4} = \frac{mole of H_{2}SO_{4}}{molarity of H_{2}SO_{4}}\)
= \(\frac{1 mole}{0.5 M} = 2cm^{3} H_{2}SO_{4}\)
The balanced equation is given as:
\(H_{2}SO_{4} + 2NaOH \to Na_{2}SO_{4} + 2H_{2}O\)
Molarity of \(NaOH = 0.1M\)
Volume of \(NaOH = 20cm^{3}\)
Molarity of \(H_{2}SO_{4} = 0.5M\)
mol NaOH = \(0.1 \times 20cm^{3} = 2molNaOH\)
⇒ \(2mol NaOH \times \frac{1mol H_{2}SO_{4}}{2mol NaOH} = 1 mol H_{2}SO_{4}\)
\(volume of H_{2}SO_{4} = \frac{mole of H_{2}SO_{4}}{molarity of H_{2}SO_{4}}\)
= \(\frac{1 mole}{0.5 M} = 2cm^{3} H_{2}SO_{4}\)