a 1.3x10-6 mol sample of Sr(OH)2 is dissolved in water to make up 25 mL os solution. what is the pH of the solution at 25 C

10.02

Explanation:

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

Mole of Sr(OH)₂ = 1.3×10¯⁶ mol

Volume of solution = 25 mL

pH of solution =?

Next, we shall convert 25 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

25 mL = 25 mL × 1 L / 1000 mL

25 mL = 0.025 L

Thus, 25 mL is equivalent to 0.025 L.

Next, we shall determine the molarity of Sr(OH)₂. This can be obtained as follow:

Mole of Sr(OH)₂ = 1.3×10¯⁶ mol

Volume of solution = 0.025 L

Molarity of Sr(OH)₂ =?

Molarity = mole / Volume

Molarity of Sr(OH)₂ = 1.3×10¯⁶ / 0.025

Molarity of Sr(OH)₂ = 5.2×10¯⁵ M

Next, we shall determine concentration of the hydroxide ion in the solution.

This can be obtained as follow:

Sr(OH)₂ (aq) <=> Sr²⁺ + 2OH¯

From the balanced equation above,

1 mole of Sr(OH)₂ produce 2 moles of OH¯.

Therefore, 5.2×10¯⁵ M Sr(OH)₂ will produce = 5.2×10¯⁵ × 2 = 1.04×10¯⁴ M OH¯

Thus, the concentration of Hydroxide ion, [OH¯] in the solution is 1.04×10¯⁴ M

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

Concentration of Hydroxide ion [OH¯] = 1.04×10¯⁴ M

pOH =?

pOH = –Log [OH¯]

pOH = –Log 1.04×10¯⁴

pOH = 3.98

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

pOH = 3.98

pH =?

pH + pOH = 14

pH + 3.98 = 14

Collect like terms

pH = 14 – 3.98

pH = 10.02

Therefore, the pH of the solution is 10.02

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