Question

A 100 mg of HNO$_3$ (strong acid) is added to water, bringing the final volume to 1.0 liter. Consider the atomic weights of H, N, and O, as 1 g/mol, 14 g/mol, and 16 g/mol, respectively. The final pH of this water is (Ignore the dissociation of water.)

• A. 2.8
• B. 6.5
• C. 3.8
• D. 8.5

Hint:

For strong acids, the pH can be calculated directly from the concentration of \( \text{H}^+ \) ions. Use \( \text{pH} = -\log[\text{H}^+] \).

Answer 1

The Correct Option is A

First, we calculate the number of moles of HNO$_3$. Given that the mass of HNO$_3$ is 100 mg = 0.1 g and the molecular weight of HNO$_3$ is 63 g/mol, the moles of HNO$_3$ are: \[ \text{Moles of HNO}_3 = \frac{\text{Mass}}{\text{Molecular weight}} = \frac{0.1 \, \text{g}}{63 \, \text{g/mol}} = 1.587 \times 10^{-3} \, \text{mol}. \] Since HNO$_3$ is a strong acid, it dissociates completely in water to give an equivalent concentration of \( \text{H}^+ \) ions. The final concentration of \( \text{H}^+ \) is: \[ [\text{H}^+] = \frac{\text{Moles of HNO}_3}{\text{Volume}} = \frac{1.587 \times 10^{-3} \, \text{mol}}{1.0 \, \text{L}} = 1.587 \times 10^{-3} \, \text{M}. \] The pH is calculated as: \[ \text{pH} = -\log[\text{H}^+] = -\log(1.587 \times 10^{-3}) = 2.8. \] Thus, the final pH is 2.8.