## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 4 - Exercises - Page 191: 114

1.67 L

#### Work Step by Step

1. Calculate the total amount of moles of $HCl$ and $H_2SO_4$. $$500.0 \space mL \times \frac{1 \space L}{1000 \space mL} \times \frac{0.100 \space mol \space HCl}{1 \space L} = 0.0500 \space mol \space HCl$$ $$500.0 \space mL \times \frac{1 \space L}{1000 \space mL} \times \frac{0.200 \space mol \space H_2SO_4}{1 \space L} = 0.100 \space mol \space H_2SO_4$$ 2. $HCl$ only has 1 $H^+$, therefore, each HCl will reach with 1 KOH. But, $H_2SO_4$ has 2, so each $H_2SO_4$ will react with 2 $KOH$. $$0.0500 \space mol \space HCl \times \frac{1 \space mol \space KOH}{1 \space mol \space HCl} = 0.0500 \space mol \space KOH$$ $$0.100 \space mol \space H_2SO_4 \times \frac{2 \space mol \space KOH}{1 \space mol \space H_2SO_4} = 0.200 \space mol \space KOH$$. 3. Determine the total amount: $$Total = 0.0500 + 0.200 = 0.250 \space mol \space KOH$$ 4. Find the necessary volume: $$0.250 \space mol \space KOH \times \frac{1 \space L}{0.150 \space mol \space KOH} = 1.67 \space L$$

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