8. Calculate (H^+), (OH^-), pOH and the pH for a 0.00024 M solution of calcium hydroxide. Must show work!

1. [H⁺] = 2.0×10¯¹¹ M

2. [OH¯] = 4.8×10¯⁴ M

3. pOH = 3.3

4. pH = 10.7

Explanation:

From the question given above, the following data were obtained obtained:

Molarity of Ca(OH)₂ = 0.00024 M

We'll begin by calculating the concentration of the hydroxide ion [OH¯]. This can be obtained as follow:

Ca(OH)₂ (aq) —> Ca²⁺ + 2OH¯

From the balanced equation above,

1 mole of Ca(OH)₂ produced 2 moles of OH¯.

Therefore, 0.00024 M Ca(OH)₂ will produce = 2 × 0.00024 = 4.8×10¯⁴ M OH¯

Thus, the concentration of the hydroxide ion [OH¯] is 4.8×10¯⁴ M

Next, we shall determine the pOH of the solution. This can be obtained as follow:

Concentration of the hydroxide ion [OH¯] = 4.8×10¯⁴ M

pOH =?

pOH = –Log [OH¯]

pOH = –Log 4.8×10¯⁴

pOH = 3.3

Next, we shall determine the pH of the solution. This can be obtained as follow:

pOH = 3.3

pH =?

pH + pOH = 14

pH + 3.3 = 14

Collect like terms

pH = 14 – 3.3

pH = 10.7

Finally, we shall determine the concentration of hydrogen ion [H⁺]. This can be obtained as follow:

pH = 10.7

Concentration of hydrogen ion [H⁺] =?

pH = –Log [H⁺]

10.7 = –Log [H⁺]

Divide both side by –1

–10.7 = Log [H⁺]

Take the antilog of –10.7

[H⁺] = Antilog (–10.7)

[H⁺] = 2.0×10¯¹¹ M

SUMMARY:

1. [H⁺] = 2.0×10¯¹¹ M

2. [OH¯] = 4.8×10¯⁴ M

3. pOH = 3.3

4. pH = 10.7

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