Question

Helpp! 100 POINTS!

What is the hydroxide ion concentration and the pH of 0.035 M Sr(OH)2 at 25 °C?

Answer 1

Strontium hydroxide, Sr(OH)2, is a strong base that dissociates completely in water to produce two hydroxide ions for every one formula unit of strontium hydroxide:

Sr(OH)2 (s) → Sr2+ (aq) + 2 OH- (aq)

Thus, the concentration of hydroxide ions in a 0.035 M solution of Sr(OH)2 is:

[OH-] = 2 × 0.035 M = 0.070 M

To find the pH of this solution, we can use the relationship between pH and the concentration of hydroxide ions:

pH = -log[H+]

Since the solution is basic, we know that the concentration of hydrogen ions is very small compared to the concentration of hydroxide ions, and we can assume that the concentration of water is constant at 55.5 M:

Kw = [H+][OH-] = 1.0 × 10^-14

Solving for [H+] gives:

[H+] = Kw / [OH-] = 1.0 × 10^-14 / 0.070 = 1.43 × 10^-13 M

Taking the negative logarithm of this value gives the pH:

pH = -log[H+] = -log(1.43 × 10^-13) = 12.85

Therefore, the hydroxide ion concentration of the 0.035 M Sr(OH)2 solution is 0.070 M, and the pH of the solution is 12.85 at 25 °C.

Answer 2

So the pH of a 0.035 M Sr(OH)2 solution at 25 °C is approximately 12.85.

Step-by-step explanation:

The dissociation of Sr(OH)2 in water can be represented as:

Sr(OH)2 (s) → Sr2+ (aq) + 2OH- (aq)

Since each formula unit of Sr(OH)2 produces two hydroxide ions, the hydroxide ion concentration ([OH-]) can be calculated as follows:

[OH-] = 2 × 0.035 M = 0.07 M

To calculate the pH, we can use the following equation:

pH = 14 - pOH

where pOH is the negative logarithm of the hydroxide ion concentration:

pOH = -log[OH-] = -log(0.07) = 1.15

Therefore,

pH = 14 - pOH = 14 - 1.15 = 12.85

So the pH of a 0.035 M Sr(OH)2 solution at 25 °C is approximately 12.85.