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:
$$pH = -log[H_3O^+] = -log( 0.105 ) = 0.979 $$
