Following five solutions of KOH were prepared as: first, 0.1 mole in 1 L; second, 0.2 mole in 2 L; t — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Observe the molarity of each individual solution: $0.1/1 = 0.1\text{ M}$, $0.2/2 = 0.1\text{ M}$, $0.3/3 = 0.1\text{ M}$, and so on. Since every individual solution has a KOH concentration of $0.1\text{ M}$, mixing any amount of them together will produce a final solution that is still $0.1\text{ M}$ KOH. Thus, $[\text{OH}^-] = 0.1\text{ M}$. The pOH is $-\log(0.1) = 1$, and the pH is $14 - 1 = 13$.