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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: paquest on February 29, 2008, 03:14:51 PM

Title: Aqueous-Percipatate Reaction Question
Post by: paquest on February 29, 2008, 03:14:51 PM

2 questions very similar.  Will give details, then what I have come up with so far.
If you can assist me in where the flaw is I would appreciate it.

Q1: An acid solution Molarity (M) is .12 in HCL & .21M in H2SO4.
What Volume(V) of .15 M KOH (Base) is needed to add .5L of acidic solution to neutralize all the acid?
My Work:
Equation I THINK:   3KOH + HCL + H2SO4 -> K2SO4 + KCL+H20
KOH + HCL -> KCL + H2O      2KOH + H2SO4 -> K2SO4 + 2 H2O

(M of acid x Vol of acid) x (mole ratio of KOH & HCL) = (M of KOH base X Vol base)
Vol H2SO4 = (1L/.15 M KOH) x (2 mole KOH/1 mole H2SO4) x (.21 mole H2SO4/1L x .5L)
Vol HCl = (1L/.15 M KOH) x (1 mole KOH/1 mole HCl) x (.12 mole HCl/1L x .5L)
Add Vol H2SO4 + Vol HCl   or 2.8 + .4 = 3.2L
also tried 3mole KOH to 1 mole H2SO4  which came to 2.1 + .4 = 2.5L
Also tried 3.6L and 2.8L but ALL are WRONG   ANY SUGGESTIONS?

Q2: Phosphates form precipitate w/ Ca & Mg to make water soft. Solution is .05 M in Calcium Chloride and .095 M in Magnesium Nitrate.  What MASS of Sodium Phosphate (g) would be added to 1 Liter (L) of solution to get rid of ALL hard water ions?

My work: Molarity = .05 moles CaCl/1 L   .095 moles MgNO3/ 1 L    ? grams Na2PO4
(M of acid x Vol of acid) x (mole ratio) = (M of base X Vol base)
But the problem above is not working so I don't know if I need another approach for this one.
Thanks for any assistance you can give me.

Title: Re: Aqueous-Percipatate Reaction Question
Post by: Borek on February 29, 2008, 04:36:17 PM

Q1 - start treating both neutralizations separately. Alternatively, start calculating total H⁺ concentration - and then neutralize this total.

Q2 - same thing. Reactions are separate, even if they occur simultaneously.

Title: Re: Aqueous-Percipatate Reaction Question
Post by: AWK on March 03, 2008, 06:49:55 AM

Q2
Ca and Mg cations react identically with phospate, hence having molar concentrations you can add them during stoichiometry, ie
 you have 0.050 + 0.095 = 0.145 M (Ca+Mg)
3Ca₂₊ + 2PO₄^3- = (Ca,Mg)₃(PO4)₂

hence you need 2x0.145/3=0.097 mole of anhydrous phosphate (which is about 16 g - crude approximation!)

Sodium phosphate is Na₃PO₄

Note this is shortcut for advanced students. You need rather Borek hint