300 ml of an aqueous solution of NaOH with pH value of 10 is mixed with 200 ml of an aqueous solution of HCl with a pH value of 4. What will be the pH of the resultant solution at room temperature?

Step-by-step solution

For the NaOH solution, pH = 10 , so pOH = 14 - 10 = 4 . [ OH ⁻] = 10⁻⁴ M . Moles of OH ⁻ = 300 10⁻³ L 10⁻⁴ mol/L = 3 10⁻⁵ mol . For the HCl solution, pH = 4 , so [ H ⁺] = 10⁻⁴ M . Moles of H ⁺ = 200 10⁻³ L 10⁻⁴ mol/L = 2 10⁻⁵ mol . Since OH ⁻ reacts with H ⁺ to form water, the remaining moles of OH ⁻ are 3 10⁻⁵ - 2 10⁻⁵ = 1 10⁻⁵ mol . Total volume of the mixture = 300 ml + 200 ml = 500 ml = 0.5 L . Resultant concentration of OH ⁻ = 1 10⁻⁵ mol 0.5 L = 2 10⁻⁵ M . pOH = - (2 10⁻⁵) = 5 - 2 = 5 - 0.301 = 4.699 . pH = 14