What is the pH of a solution prepared by dissolving 0.140 g of potassium hydroxide in sufficient pure water to prepare 250.0 ml of solution

pH = 12

Explanation:

Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water by giving OH⁻ anions as follows:

KOH⇒ K⁺ + OH⁻

Since dissociation is complete, it is assumed that the concentration of OH⁻ is equal to the initial concentration of KOH:

[OH⁻] = [KOH]

In order to find the initial concentration of KOH, we have to divide the mass (0.140 g) into the molecular weight of KOH (Mw):

Mw (KOH) = K + O + H = 39 g/mol + 16 g/mol + 1 g/mol = 56 g/mol

moles KOH: mass/Mw = 0.140 g / (56 g/mol) = 2.5 x 10⁻³ moles

The molality of the solution is the number of moles of KOH per liter of solution:

V = 250.0 ml x 1 L/1000 ml = 0.250 L

M = (2.5 x 10⁻³moles) / (0.250 L) = 0.01 M

Now, we calculate pOH:

pOH = - log [OH⁻] = - log [KOH] = - log (0.01) = 2

Finally, we calculate pH from pOH:

pH + pOH = 14

⇒pH = 14 - pOH = 14 - 2 = 12