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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Answer:
Group B elements
Explanation:
Group B metals are referred to as transition metals. They are located in the middle of the periodic table between Group IIA and Group IIIA. They provide transition between the "base formers" on the left and "acid formers" on the right.
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Explanation: hope this helps.
