Answer to Question #109046 in Chemistry for vai

Question

in an experiment,50.0 cm 3 of magnesium chloride were titrated with 1.00 mol/dm3 sodium hydroxide. calculate the concentration of magnesium chloride

Answer 1

Not a complete data set is given in the task

Solution:

Magnesium chloride react with sodium hydroxide to produce magnesium hydroxide and sodium chloride:

MgCl₂ + 2NaOH → Mg(OH)₂ + 2NaCl

According to the chemical equation: n(MgCl₂) = n(NaOH)/2.

n(MgC₂) = C(MgCl₂)*V(MgCl₂), where C(MgCl₂) is the concentration of magnesium chloride; V(MgCl₂) is the volume of the magnesium chloride solution.

n(NaOH) = C(NaOH) * V(NaOH), where C(NaOH) is the concentration of sodium hydroxide; V(NaOH) is the volume of the sodium hydroxide solution.

Then,

C(MgCl₂)*V(MgCl₂) = C(NaOH)*V(NaOH)/2;

C(MgCl₂) = [C(NaOH) * V(NaOH)] / [V(MgCl₂) * 2]

C(MgCl₂) = [ 1.00 mol/dm³ * V(NaOH) ] / [ 0.05 dm³ * 2 ] = 10 * V(NaOH)

C(MgCl₂) = 10 * V(NaOH).

In the condition of the task is not given the volume of the sodium hydroxide (NaOH) solution that was used in the titration.

Answer: C(MgCl₂) = 10 * V(solution of NaOH).