Q.Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.

The key idea here is calculating the pH of a mixture of strong acids. Since both are strong acids, they dissociate completely, and the hydrogen ion concentration determines the pH.

1. First, determine the hydrogen ion concentration for each solution from its pH:

   For solution A: $[H^+]_A = 10^{-pH_A} = 10^{-6} \text{ M}$

   For solution B: $[H^+]_B = 10^{-pH_B} = 10^{-4} \text{ M}$

2. Assume equal volumes, say $V$ litres, for each solution. Calculate the total moles of $H^+$ ions:

   Moles of $H^+$ from A $= [H^+]_A \times V = 10^{-6} V$ moles

   Moles of $H^+$ from B $= [H^+]_B \times V = 10^{-4} V$ moles

   Total moles of $H^+ = (10^{-6} V + 10^{-4} V) = V(10^{-6} + 10^{-4}) = V(0.000001 + 0.0001) = 1.01 \times 10^{-4} V$ moles

3. The total volume of the mixture is $V + V = 2V$ litres. Calculate the final hydrogen ion concentration: …

The pH scale is a logarithmic measure of the hydrogen ion concentration ($[H^+]$) in a solution, defined as $pH = -\log_{10}[H^+]$. This means that a change of one pH unit represents a tenfold change in $[H^+]$. Consequently, pH values cannot be simply averaged when mixing solutions.

When two solutions are mixed, the total number of moles of $H^+$ ions in the mixture is the sum of the moles of $H^+$ ions from each individual solution. The total volume is the sum of the individual volumes. The new concentration of $H^+$ ions is then calculated by dividing the total moles of $H^+$ by the total volume. Finally, this new concentration is converted back to pH.

For strong acids, like the ones in this problem, they dissociate completely in water. This means that if a strong acid has a concentration of $C$ M, the concentration of $H^+$ ions it contributes will also be $C$ M.

Let's proceed with the calculation step-by-step.

1. Determine the hydrogen ion concentration ($[H^+]$) for each solution.

   The relationship between pH and $[H^+]$ is given by $[H^+] = 10^{-pH}$.

   • For solution A, $pH_A = 6$:

     $[H^+]_A = 10^{-6} \text{ M}$

   • For solution B, $pH_B = 4$:

     $[H^+]_B = 10^{-4} \text{ M}$

   Watch out

   It is a common mistake to directly average the pH values. This is incorrect because pH is a logarithmic scale. Always convert pH to concentration before performing calculations involving mixing.

2. Calculate the moles of $H^+$ in each solution.

   The problem states that equal volumes of solutions A and B are mixed. Let's assume the volume of each solution is $V$ litres.

   Moles of $H^+$ = Concentration $\times$ Volume

   • Moles of $H^+$ in solution A ($n_A$):

     $n_A = [H^+]_A \times V = 10^{-6} \times V \text{ moles}$

   • Moles of $H^+$ in solution B ($n_B$):

     $n_B = [H^+]_B \times V = 10^{-4} \times V \text{ moles}$

3. Calculate the total moles of $H^+$ after mixing.

   The total moles of $H^+$ ($n_{total}$) in the mixture will be the sum of the moles from each solution:

   $n_{total} = n_A + n_B = (10^{-6} \times V) + (10^{-4} \times V)$

   $n_{total} = V (10^{-6} + 10^{-4})$

   $n_{total} = V (0.000001 + 0.0001)$

   $n_{total} = V (0.000101) \text{ moles}$

4. Calculate the total volume after mixing.

   Since the volume of each solution is $V$ litres, the total volume ($V_{total}$) after mixing will be:

   $V_{total} = V_A + V_B = V + V = 2V \text{ litres}$

5. Calculate the new hydrogen ion concentration ($[H^+]_{new}$) in the mixture.

   The new concentration is the total moles of $H^+$ divided by the total volume:

   $[H^+]_{new} = \frac{n_{total}}{V_{total}} = \frac{V (0.000101)}{2V}$

   $[H^+]_{new} = \frac{0.000101}{2}$

   $[H^+]_{new} = 0.0000505 \text{ M}$

   $[H^+]_{new} = 5.05 \times 10^{-5} \text{ M}$ …