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To 0.360 l of 0.160 m nh3 is added 0.120 m mgcl2. how many grams of (nh4)2so4 should be present to prevent the precipitation of mg(oh)2? the ksp of mg(oh)2 is 1.8x10-11.

Respuesta :

taskmasters
first calculate the moles of every compound used

n NH3 = (0.160 mol /L )(0 .360 L) = 0.058 moles NH3
which is equal to moles NH4OH
0.058 moles NH4OH.

  n MgCl2 = (0.12 mol /L)( 0.10 liters = .012 moles MgCl2

reaction 2 NH3.H20 or 2 NH4OH + MgCl2 --> Mg(OH)2(s) + 2 NH4+ + 2Cl-

Molarity of Mg++ =(0.012 moles Mg+) / 0.46 liters = 0.026 Molar Mg++

Molarity OH= (0.058 moles OH-)/ 0.46 liters = 0.126 molar OH-

the limiting reactant is Mg ++ because it is lesser than the molarity of OH-

Now the challenge is to drive the OH-) concentration down so low that Mg(OH)2 will not precipitate out.

Ksp = 1.8 x 10^-11 = (Mg)(OH-)^2

  Mg++ concentration to be .026 Molar, so let X = the (OH-) concentration

1.8 x l0^-11 = (0.026)(X)^2

(X)^2 = ( 1.8 x 10^-11) / (0.026)

X^2 = 6.92 x 10^-10
X = 2.63 x l0^-5 moles (OH-)/ L
this is the concentration where solid will form

so we need to lower the (OH-) which is 0.126 molar OH- down to 2.63 x 10^-5 molar OH- by adding NH4+ ions.

(0.126 moles/liter 0H-) - (2.63 x l0^-5 molar OH- ) = 0.123 moles per liter

OH- to be neutralized by adding NH4+

since the mole ratio is 1 : 1 then  0.123 moles per liter NH4+ concentration to neutralize the  0.123 moles of OH- in solution.


so to prevent the precipitation of mg(oh)2

0.123 - 0.058 = 0.065 Molar NH4+ is needed


mahima6824soni

The grams of [tex](NH_4)2SO_4[/tex] present to prevent the precipitation of [tex]Mg(OH)_2[/tex] is 0.065.

What is precipitation?

To convert a miscible liquid into immiscible solid is called precipitation.

The reaction is

[tex]\rm 2 NH_4OH + MgCl_2 -- > Mg(OH)_2(s) + 2 NH_4+ + 2Cl-[/tex]

Step1 : calculate the moles

[tex]\rm Number\;of \;moles= \dfrac{mass}{molar\;mass}\\\rm Number\;of \;moles\;of\; NH_3 = \dfrac{0.160 mol /L}{0 .360 L}=0.058 mol NH_3[/tex]

[tex]NH_4OH =[/tex] 0.058 mol

[tex]\rm Number\;of \;moles\;of\; MgCl_2 = \dfrac{0.12 mol /L}{0.10 l}=0.12\; mol\; MgCl_2[/tex]

Step2 : calculating the molarity

[tex]\rm M = \dfrac{n}{V}\\Molarity\;of Mg++ = \dfrac{0.012 moles Mg+}{0.46\; l} = 0.026 \\\\\rm Molarity\;of \;OH = \dfrac{0.058 m}{0.46 l} = 0.126[/tex]

The Mg++ has lower molarity, so it is a limiting reagent.

Step 3: Lowering the concentration of OH-

ksp =[tex]1.8 \times 10^-^1^1 = (Mg)(OH^-)^2[/tex]

[tex]1.8 \times l0^-^1^1 = (0.026)(X)^2\\(X)^2 = \dfrac{1.8 \times 10^-^1^1}{ (0.026)} \\X^2 = 6.92 \times 10^-^1^0\\\\X = 2.63 \times l0^-^5 moles (OH-)/ L[/tex]

The concentration is solid now.

Now lowering the molar mass of OH-

[tex]\rm (0.126 mol/l\; OH-) - (2.63 \times l0^-^5 M\; OH^- ) = 0.123 mol/l[/tex]

Neutralize the OH- by adding NH4+

The mole ratio is 1:1

To stop the precipitation of Mg(OH)2

0.123 - 0.058 = 0.065 M

Thus, the [tex](NH_4)2SO_4[/tex] needed is 0.065.

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