A chemist prepared a solution of NaOH by completely dissolving 55.90 grams of solid NaOH in 3.600 liters of water at room temperature. What was the pH of the solution that the chemist prepared, to the nearest thousandth?

13.59

Explanation:

Step 1:

We'll begin by writing out the data from the question.

Mass of NaOH = 55.90g

Volume = 3.6 L

pH =?

Step 2:

Determination of the number of mole NaOH.

Mass of NaOH = 55.90g

Molar Mass of NaOH = 23 + 16 +1 = 40g/mol

Number of mole = Mass/Molar Mass

Number of mole of NaOH = 55.90/40

Number of mole of NaOH = 1.3975 mole

Step 3:

Determination of the molarity of NaOH.

Molarity = mole /Volume

Mole of NaOH = 1.3975 mole

Volume = 3.6 L

Molarity of NaOH = 1.3975/3.6

Molarity of NaOH = 0.388 M

Step 4:

Determination of the concentration of OH- in 0.388 M NaOH

In solution, NaOH will dissociate as follow:

NaOH —> Na+ + OH-

From the equation above,

1 mole of NaOH produced 1 mole of OH-

Therefore, 0.388 M of NaOH will also produce 0.388 M of OH-

The concentration of OH- is 0.388 M

Step 5:

Determination of the pOH of the solution.

pOH = - Log [OH-]

Concentration of OH-, [OH-] = 0.388 M

pOH = - Log [OH-]

pOH = - Log 0.388

pOH = 0.411

Step 6:

Determination of the pH of the solution.

pH + pOH = 14

pOH = 0.411

pH + 0.411 = 14

pH = 14 - 0.411

pH = 13.589

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