Questions: A chemist dissolves 390 mg of pure nitric acid in enough water to make up 240 mL of solution. Calculate the pH of the solution. Round your answer to 3 significant decimal places.

A chemist dissolves 390 mg of pure nitric acid in enough water to make up 240 mL of solution. Calculate the pH of the solution. Round your answer to 3 significant decimal places.

Transcript text: A chemist dissolves $390 . \mathrm{mg}$ of pure nitric acid in enough water to make up $240 . \mathrm{mL}$ of solution. Calculate the pH of the solution. Round your answer to 3 significant decimal places. $\square$

Solution

Solution Steps

Step 1: Determine the Molar Mass of Nitric Acid

Nitric acid ($\text{HNO}_3$) has a molar mass calculated as follows:

Hydrogen (H): $1.008 \, \text{g/mol}$
Nitrogen (N): $14.01 \, \text{g/mol}$
Oxygen (O): $16.00 \, \text{g/mol} \times 3 = 48.00 \, \text{g/mol}$

Adding these together gives the molar mass of $\text{HNO}_3$: \[ 1.008 + 14.01 + 48.00 = 63.018 \, \text{g/mol} \]

Step 2: Calculate the Moles of Nitric Acid

Convert the mass of nitric acid from milligrams to grams: \[ 390 \, \text{mg} = 0.390 \, \text{g} \]

Calculate the moles of $\text{HNO}_3$ using its molar mass: \[ \text{moles of } \text{HNO}_3 = \frac{0.390 \, \text{g}}{63.018 \, \text{g/mol}} = 0.006188 \, \text{mol} \]

Step 3: Calculate the Concentration of Nitric Acid

Convert the volume of the solution from milliliters to liters: \[ 240 \, \text{mL} = 0.240 \, \text{L} \]

Calculate the concentration of $\text{HNO}_3$ in the solution: \[ \text{Concentration} = \frac{0.006188 \, \text{mol}}{0.240 \, \text{L}} = 0.025783 \, \text{M} \]

Step 4: Calculate the pH of the Solution

Since nitric acid is a strong acid, it dissociates completely in water. Therefore, the concentration of $\text{H}^+$ ions is equal to the concentration of $\text{HNO}_3$.

Calculate the pH: \[ \text{pH} = -\log_{10}(0.025783) = 1.588 \]