What is the pOH of a solution prepared by dissolving 0.655g of KOH(s) in 3.00 L of water

2.4

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass of KOH = 0.655g

Volume of H2O = 3L

pOH =..?

Step 2:

Determination of the number of mole of KOH. This is illustrated below:

Mass of KOH = 0.655g

Molar mass of KOH = 39 + 16 + 1 = 56g/mol

Number of mole KOH =..?

Mole = Mass /Molar Mass

Number of mole of KOH = 0.655/56

Number of mole of KOH = 0.012 mol

Step 3:

Determination of the molarity of KOH.

The molarity of KOH can be obtained as follow:

Mole of KOH = 0.012 mol

Volume of water = 3L

Molarity =..?

Molarity = mole /Volume

Molarity of KOH = 0.012/3

Molarity of KOH = 4x10^–3 M

Step 4:

Determination of the concentration of Hydroxide ion, [OH-] in 4x10^–3 M KOH.

This can be obtained as follow:

The dissociation equation of KOH is given below:

KOH —> K+ + OH-

From the balanced equation above,

1 mole of KOH produced 1 mole of OH-.

Therefore, 4x10^–3 M KOH will also produce 4x10^–3 M OH-

The concentration of Hydroxide ion, [OH-] = 4x10^–3 M

Step 5:

Determination of the pOH of the solution.

pOH = –Log [OH-]

[OH-] = 4x10^–3 M

pOH = –Log 4x10^–3

pOH = 2.4

Therefore, the pOH of the solution is 2.4

i think that the answer is c