Calculate the pHpH of a 0.10 MM solution of barium hydroxide, Ba(OH)2Ba(OH)2. Express your answer numerically using two decimal places.

13.301

Explanation:

To calculate the pH of the solution, we must obtain the pOH of the solution as illustrated below:

The dissociation equation is given below

Ba(OH)2 <==> Ba^2+ + 2OH^-

Since Ba(OH)2 dissociate to produce 2moles of OH^-, the concentration of OH^- = 2x0.1 = 0.2M

pOH = - Log[OH^-]

pOH = - Log 0.2

pOH = 0.699

But

pH + pOH = 14

pH = 14 — pOH

pH = 14 — 0.699

pH = 13.301

The pH of this barium hydroxide solution is 13.30

Explanation:

Step 1: Data given

Concentration Ba(OH)2 = 0.10 M

Step 2: Calculate [OH-]

Ba(OH)2 ⇒ Ba^2+ + 2OH-

[OH-] = 2*0.10 M

[OH-] = 0.20 M

Step 3: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.20)

pOH = 0.70

Step 4: Calculate pH

pH + pOH = 14

pH = 14 -pOH

pH = 14 - 0.70

pH = 13.30

The pH of this barium hydroxide solution is 13.30