Question

A solution contains Cr3+ and Mg2+. The addition of 1.00 L of 1.51 M NaF solution is required to cause the complete preci...


Dnyanesh Vernekar - Chemistry

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Transcribed: A solution contains Cr3+ and Mg2+. The addition of 1.00 L of 1.51 M NaF solution is required to cause the complete precipitation of these ions as CrF3(s) and MGF2(s). The total mass of the precipitate is 50.1 g. Find the mass of Cr3+ in the original solution.

Answer

The balanced equations for the precipitation of Cr3+ and Mg2+ using NaF can be written as below

Cr3+ + 3NaF CrF+ 3Na+

Mg2+ + 2NaF MgF+ 2Na+

The molecular weight (or molar mass) of CrF= (52+3x19)= 109.0 g

The molecular weight (or molar mass) of MgF2 = (24.3 +2x19) = 62.3 g

Molarity (M) of the solution is defined as the number of moles of the solute present (dissolved) in 1 liter of the solution.

The concentration of NaF is given as 1.51 M.

To get mol of F

mol of F1L=1.51Mmol of F =1.51M x 1L= 1.51

Thus 1.51 moles of F- ions are used to precipitate.

Also we know that,

mass of CrF3 + mass of MgF2 = total mass of precipitate 

if x = mass of CrF3 and y =mass of MgF2

then x + y= 50.1     .......     equation 1

From the balanced equation given in Step 1 

we require 3 mol of F to get 1 mol of CrF3 and 2 mol of F to get 1 mol of MgF2

Hence, we can represent this in equation form as

3 (mol of CrF3) + 2 (mol of MgF2) = Mol of F

By definition 

Mol = weight/molecular weight

Hence,

3×mass (CrF3)Molecular weight (CrF3)+2×mass (MgF2)Molecular weight (MgF2)=Mol of F

substituting the values we get

3x109+2y62.3=1.5      ............... equation  2

 

simplifying  Equation 2, we get

2y=62.3×(1.5-3x109)2y=62.3×(163.5-3x109)y=0.285 ×(163.5-3x)y=46.76-0.856x

substituting this value of y in equation 1, we get

x+46.76-0.857x=50.10.144x=50.1-46.760.144x=3.34x=23.19g

Thus 23.19g of CrF3 is precipitated.

Mol = weight/molecular weight

hence, mol of Cr originally present = 23.19/109.0

Mol of Cr = 0.213 moles

number of moles = weight/ molecular weight

molecular weight of Cr is 52

Hence mass of Cr3+ in the original solution = 0.213 x52 = 11.08g

The mass of Cr3+ is 11.08g in the original solution

 

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