Which volume of 0.5 M H 2SO4 will exactly neutralize 20 cm 3 of 0.1 M NaOH solution?
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Correct Answer: Option A
Explanation:
\(\frac{C_AV_A}{C_bV_b} = \frac{1}{2}\)
\(frac{0.5 \times V_A}{0.1 \times 20} = \frac{1}{2}\)
\(V_A = \frac{0.1 \times 20}{2 \times 0.5} = \frac{1.0}{0.5)\)
= 2cm2
\(\frac{C_AV_A}{C_bV_b} = \frac{1}{2}\)
\(frac{0.5 \times V_A}{0.1 \times 20} = \frac{1}{2}\)
\(V_A = \frac{0.1 \times 20}{2 \times 0.5} = \frac{1.0}{0.5)\)
= 2cm2