A 1. 3×10−6mol sample of sr(oh)2 is dissolved in water to make up 25. 0 ml of solution. what is the ph of the solution at 25. 0∘

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1 year ago

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VMariaS [17] · Answer 1 · 2023-08-20T02:42:35+03:00

The pH of the solution at 25 degree celsius of 1.3 × 10⁻⁶ moles of a sample of Sr(OH)₂ is 10.02.

<h3>How do we calculate pH?</h3>

The pH of any solution gives an idea about the acidic and basic nature of the solution and the equation of pH will be represented as:

pH + pOH = 14

Given that,

Moles of Sr(OH)₂ = 1.3 × 10⁻⁶ mol

Volume of solution = 25mL = 0.025L

The concentration of Sr(OH)₂ in terms of molarity = 1.3×10⁻⁶/0.025

= 5.2×10¯⁵M

Dissociation of Sr(OH)₂ takes place as:

Sr(OH)₂ → Sr²⁺ + 2OH⁻

From the stoichiometry of the reaction 1 mole of Sr(OH)₂ produces 2 moles of OH⁻.

Given that the base is a strong base and that it entirely dissociates into its ions, the hydroxide ion concentration is 5.2×10¯⁵×2 = 1.04×10¯⁴ M.

pOH = -log[OH⁻]

pOH = -log(1.04×10¯⁴)

pOH = 3.98

Now we put this value on the first equation we get,

pH = 14 - 3.98 = 10.02

Therefore, the value of pOH is 10.02.

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