Answer:
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
