## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 57

#### Answer

\begin{vmatrix} & [H_3O^+] & [OH^-] & pH \\ a. & 0.25 & 4.0 \times 10^{-14} & 0.60 \\ b. & 0.015 & 6.7 \times 10^{-13} & 1.82 \\ c. & 0.072 & 1.4 \times 10^{-13} & 1.14 \\ d. & 0.150 & 9.5 \times 10^{-14} & 0.979 \end{vmatrix}

#### Work Step by Step

a. 1. Since $HCl$ is a strong acid, all its concentration is going to produce $H_3O^+$: $$[H_3O^+] = [ HCl ] = 0.25 \space M$$ 2. Calculate the $[OH^-]$: $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.25 } = 4.0 \times 10^{-14} \space M$$ 3. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.25 ) = 0.60$$ b. 1. Since $HNO_3$ is a strong acid, all its concentration is going to produce $H_3O^+$: $$[H_3O^+] = [ HNO_3 ] = 0.015 \space M$$ 2. Calculate the $[OH^-]$: $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.015 } = 6.7 \times 10^{-13} \space M$$ 3. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.015 ) = 1.82$$ c. 1. Since $HBr$ and $HNO_3$ are both strong acids, all their concentration is going to produce $H_3O^+$: $$[H_3O^+] = [ HBr ] + [HNO_3] = 0.072 \space M$$ 2. Calculate the $[OH^-]$: $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.072 } = 1.4 \times 10^{-13} \space M$$ 3. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.072 ) = 1.14$$ d. 1. Calculate the molar mass: $HNO_3$ : ( 1.008 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 63.02 g/mol 2. Use the informations as conversion factors to find the molarity of this solution: $$\frac{ 0.655 g \space HNO_3 }{100 \space g \space solution} \times \frac{1 \space mol \space HNO_3 }{ 63.02 \space g \space HNO_3 } \times \frac{ 1.01 \space g \space solution}{1 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} = 0.105 \space M$$ 3. Since $HNO_3$ is a strong acid, all its concentration is going to produce $H_3O^+$: $$[H_3O^+] = [ HNO_3 ] = 0.105 \space M$$ 4. Calculate the $[OH^-]$: $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.105 } = 9.5 \times 10^{-14} \space M$$ 5. Calculate the pH: $$pH = -log[H_3O^+] = -log( 0.105 ) = 0.979$$

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