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andrew11 [14]
3 years ago
15

Potassium hydroxide is used to precipitate each of the cations from their respective solution. determine the minimum concentrati

on of koh required for precipitation to begin in each case.
Chemistry
1 answer:
MAXImum [283]3 years ago
8 0
This is the three cases that help to determine the minimum concentration of KOH required for precipitation 
Part a) 1.5×10^−2 M K CaCl2 
Part b) 2.3×10^−3 M Fe (NO3)2 
Part c) 2.0×10^−3 M MgBr2

a) CaCl2 + 2KOH --> Ca (OH) 2 + 2KCl Ca (OH) 2 <=> Ca^2+ + 2OH^- 
ksp = 1.5*10^-2 + x^2 
4.68*10^-6 = 1.5*10^-2 + x^2 
x= [KOH] = 0.01766 

b) Fe (NO3)2 +2 KOH--> Fe (OH)2 + 2KNO3 
Fe (OH)2 <=> Fe^2+ + 2OH^- 
ksp = 2.3*10^-3 + x^2 
4.87*10^-17 = 2.3*10^-3 + x^2 
x= 1.46*10^-7 

c) MgBr2 + KOH --> Mg (OH) 2 + 2KBr 
Mg (OH) 2 <=> Mg^2+ + 2OH^- 
ksp = 2.0*10^-3 + x^2 
2.06*10^-13 = 2.0*10^-3 + x^2 
x= 1.015*10^-5
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5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H
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We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
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Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

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Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
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I hope this helps.

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4 years ago
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