If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K.

Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Given:

Mass of KOH (m) = 0.561g

Volume(V) = 200 ml = L = 0.2L

Molar mass of KOH (M) = 56 g/mol

To calculate concentration of KOH, we apply the formula,

Concentration=

⇒[KOH] =

⇒[KOH] = 0.05 M

Now, the ionisation of KOH

⇒KOH↔ K + OH^-

At equilibrium

∴ [K] = [OH^-] = 0.05 M

To calculate [H⁺], we apply the formula:

K_w = [H⁺] [OH^-]

Where K_w is the ionic product of water which is defined as the product of molar concentration of H⁺ and OH^- ions.

Using the formula, we write

[H⁺] =

As the value of K_w is taken as 10^-14

[OH^-]= 0.05 M (calculated)

∴ [H⁺] =

⇒[H⁺] = 2.0 × 10^-13

Thus, the concentration of potassium, hydrogen, and hydroxyl ions are 0.05, 0.05 and 2.0 × 10^-13 respectively

To calculate pH, we apply the formula:

pH = –log [H⁺]

⇒pH = - log (2.0 × 10^-13)

⇒pH = 13- log2

⇒pH = 12.69

Thus, the pH is 12.69