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How many milliliters of 0. 0850 m naoh are required to titrate 25. 0 ml of 0. 0720 m hbr to the equivalence point?

Sagot :

The final answer is 29.8 mL

Molar mass HCl = 36.5 g/mol

Mol HCl in 1.85 g = 1.85 g / 36.5 g/mol = 0.05068 mol

The molarity of the HCl solution is 0.0507 mol /L

Mol HCl in 50 mL = 50 mL / 1000 mL/L * 0.0507 mol /L = 0.002535 mol

This will require 0.002535 mol NaOH

1000 mL contains 0.0850 mol

The volume that contains 0.002535 mol

Volume = 0.002535 mol / 0.0850 mol /L *1000 mL/L = 29.8 mL

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The volume of 0.0850 M NaOH is required to titrate 25.0 ml of 0.0720 M is 21.176 ml.

What are normality and molarity?

Normality is defined as the number of equivalents per litre of the solution

It is given by

Normality = number of equivalents / 1 L of solution

Molarity is defined as the number of moles of solute present in one litre of solution

Molarity = moles of solute / 1 litre of solution

Relationship between molarity and normality

Normality = Molarity × Acidity or Basicity of a salt

Here, the Acidity Of NaOH is 1 and the basicity of HBr is also 1.

Thus, Normality = Molarity

We know, N₁V₁ = N₂V₂

We can also write it as M₁V₁ = M₂V₂

V1 = [tex]\frac{M_2V_2}{M_!}[/tex]

Substitute M₁ = 0.0850 M, M₂ = 0.0720 M, V₂= 25 ml

V1 = 21.176 ml

Hence, The volume of 0. 0850 m NaOH are required to titrate 25. 0 ml of 0. 0720 m is 21.176 ml.

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