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

it's filtration, but this is an assumption so my apologies if i got this