Question

A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of...

A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added, and the resulting solution is titrated with 2.50 M solution of HCl. The indicator changes colour, signalling that the equivalence point has been reached after 17.0 mL of the hydrochloric acid solution has been added.

  • Determine the molality of this alkali metal hydroxide solution and What is the mass percentage of the metal hydroxide in the solution?

  • Determine the boiling point of water when this mass of the alkali metal hydroxide has been added. Does the boiling change significantly upon the addition of this metal hydroxide?
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Answer #1

Solution:

An Unknown Solute which is an Alkali Metal Hydroxide is dissolved in Water to Prepare a Solution.This Solution is Titrated with HCl Solution using an Acid-Base Indicator.

i) Here, 4.36g of Unknown Solute is Added.

Volume of HCl Added to the Aqueous Solution of the Solute to achieve Equivalence Point = 17ml = 0.017L.

Concentration of HCl = 2.5M

The Equivalence Point in an Acid- Base Titration is a Point where the Amount of Acid = Amount of Base in Solution.

So Number of Moles of Base Solute= Number of Moles of HCl.

Moles of Solute (n) = C×V = 2.5M× 0.017L = 0.0425moles.

Mass of Solvent(Water) = 100ml = 100g = 0.1Kg

Therefore Molality of Solution(m) = Moles of Solute÷ Mass of Solvent

m = 0.0425÷ 0.1 = 0.425 Kg^-1.

Mass of Solution = Mass of Solute + Mass of Solvent = 4.36 + 100 = 104.36g.

Therefore, Mass Percentage of the Solute (M%)= (Mass of Solute ÷Mass of Solution)×100

M% = (4.36÷104.36) ×100 = 4.18%.

ii) The Change in Boiling Point of Solution can be determined from the Molality of Solution by the following Equation:

∆Tb = Kb × m, where Kb is Molal Boiling Point Elevation Constant of Water (0.51°C/m) and m is the Molality of Solution.

Initial Temperature of Water = 100°C.

Tf is the Final Temperature of Solution after Addition of Solute

So; Tf - Ti = 0.51 × 0.425.

Tf - 100°C = 0.217.

Therefore, Tf = 0.217 + 100 = 100.217°C.

Therefore the Final Temperature of Solution after Adding the Solute is Increasing and is not a Significant Increase.

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