Calculating pH for $HNO_3$ Solution

This solution explains how to calculate the pH of a solution prepared by dissolving a specific amount of nitric acid ($HNO_3$) in water. We will determine the concentration of hydrogen ions ($[H^+]$) and then use the pH formula.

Step 1: Determine the Molarity of the $HNO_3$ Solution

Molarity is a measure of concentration, defined as the number of moles of solute per liter of solution. The formula is:

$$ Molarity (M) = \frac{\text{Moles of solute}}{\text{Volume of solution in Liters}} $$

In this problem:

• Moles of $HNO_3$ = $0.0025$ moles
• Volume of water = $250\,mL$. To use the formula, we need to convert this volume to Liters:

  $$ 250\,mL \times \frac{1\,L}{1000\,mL} = 0.250\,L $$

Now, we can calculate the molarity of the $HNO_3$ solution:

$$ M_{HNO_3} = \frac{0.0025\, \text{moles}}{0.250\, \text{L}} $$

$$ M_{HNO_3} = 0.01\, \text{mol/L} $$

So, the concentration of the nitric acid solution is $0.01\, M$.

Step 2: Find the Hydrogen Ion Concentration $[H^+]$

Nitric acid ($HNO_3$) is a strong acid. This means it dissociates completely in water into its ions, hydrogen ions ($H^+$) and nitrate ions ($NO_3^-$).

The dissociation reaction is:

$$ HNO_3(aq) \rightarrow H^+(aq) + NO_3^-(aq) $$

Because the dissociation is complete, the concentration of hydrogen ions ($[H^+]$) in the solution is equal to the initial molarity of the $HNO_3$ solution.

$$ [H^+] = M_{HNO_3} = 0.01\, M $$

Step 3: Calculate the pH

The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration:

$$ pH = -\log_{10}[H^+] $$

Substitute the value of $[H^+]$ we found:

$$ pH = -\log_{10}(0.01) $$

To make the calculation easier, we can express $0.01$ in scientific notation:

$$ 0.01 = 1 \times 10^{-2} $$

Now substitute this back into the pH formula:

$$ pH = -\log_{10}(1 \times 10^{-2}) $$

Using the properties of logarithms ($\log(a \times 10^b) = \log(a) + b$ and $\log_{10}(10^b) = b$):

$$ pH = -(-2) $$

$$ pH = 2 $$

Conclusion

The pH of the solution prepared by dissolving $0.0025$ moles of $HNO_3$ in $250\,mL$ of water is 2.0.