## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 62

#### Answer

$$Volume = 9 \times 10^{-3} \space L = 9 \space mL$$

#### Work Step by Step

1. Calculate the $[H_3O^+]$: $$[H_3O^+] = 10^{-pH} = 10^{-1.8} = 2 \times 10^{-2} \space M$$ 2. Since HCl is a strong acid, we can say that $[HCl] = [H_3O^+] = 2 \times 10^{-2} \space M$ 3. Molar mass of HCl: $HCl$ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 36.46 g/mol 4. Find the volume of the concentrated solution: $$5.00 \space L \times \frac{2 \times 10^{-2} \space mol}{1 \space L} \times \frac{36.46\space g}{1 \space mol} \times \frac{100 \space g \space solution}{36.0 \space g} \times \frac{1 \space mL \space solution}{1.179 \space g \space solution} \times \frac{1 \space L}{1000 \space mL}$$ $$Volume = 9 \times 10^{-3} \space L = 9 \space mL$$

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