Calculating Final pH After Mixing Acidic Solutions

To find the pH of the mixture, we first need to determine the concentration of hydrogen ions ($[H^+]$) in each solution and then calculate the final concentration after mixing.

Step 1: Calculate initial hydrogen ion concentrations

The relationship between pH and hydrogen ion concentration is given by the formula: $pH = -\log_{10}[H^+]$ Therefore, $[H^+] = 10^{-pH}$.

• For the first solution: $pH = 1$. $[H^+]_1 = 10^{-1} \text{ M} = 0.1 \text{ M}$
• For the second solution: $pH = 2$. $[H^+]_2 = 10^{-2} \text{ M} = 0.01 \text{ M}

Step 2: Calculate moles of H⁺ in each solution

Moles = Concentration × Volume. The volume is 50 ml, which is 0.050 L.

• Moles of $H^+$ in the first solution: $n_{H^+1} = [H^+]_1 \times V_1 = 0.1 \text{ mol/L} \times 0.050 \text{ L} = 0.005 \text{ mol}$
• Moles of $H^+$ in the second solution: $n_{H^+2} = [H^+]_2 \times V_2 = 0.01 \text{ mol/L} \times 0.050 \text{ L} = 0.0005 \text{ mol}

Step 3: Calculate total moles and final concentration

The total volume after mixing is $50 \text{ ml} + 50 \text{ ml} = 100 \text{ ml}$, which is 0.100 L.

• Total moles of $H^+$ = $n_{H^+1} + n_{H^+2} = 0.005 \text{ mol} + 0.0005 \text{ mol} = 0.0055 \text{ mol}$
• Final concentration of $H^+$: $[H^+]_{final} = \frac{\text{Total moles } H^+}{\text{Total Volume}} = \frac{0.0055 \text{ mol}}{0.100 \text{ L}} = 0.055 \text{ M}

Step 4: Calculate the final pH

Using the pH formula:

• $pH_{final} = -\log_{10}[H^+]_{final} = -\log_{10}(0.055)$
• $pH_{final} \approx 1.2596$

Rounding to two decimal places, the pH of the mixture is approximately 1.26.