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∘c?

Answers

Answer 1

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.

How do we calculate pH?

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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Related Questions

The molar solubility of cui is 2. 26 × 10-6 m in pure water. Calculate the ksp for cui.

Answers

Ksp(the solubility product constant) = [Cu⁺] [I⁻]

So, the Ksp for Cui would be:

Ksp = (2.26 × 10⁻⁶) (2.26 × 10⁻⁶) = 5.11 x 10⁻¹²

Formula used:

K = [tex]K_{sp} = [A^+]^a [B^-]^b[/tex], where

Ksp = solubility product constant

A⁺ = cation in an aquious solution

B⁻ = anion in an aqueous solution

a, b = relative concentrations of a and b

Definition

The equilibrium constant for a solid material dissolving in an aqueous solution is the SOLUBILITY PRODUCT CONSTANT, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is.

Take into account the general dissolving response (in aqueous solutions) below:

aA(s)⇔cC(aq)+dD(aq)

The molarities or concentrations of the products (cC and dD) must be multiplied in order to find the Ksp. Any product that has a coefficient in front of it must be raised to the power of that coefficient (and also multiply the concentration by that coefficient).

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