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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: bokwandu on March 18, 2008, 03:46:23 PM

Title: Using the Common Ion Effect to Remove Ions from Hard Water
Post by: bokwandu on March 18, 2008, 03:46:23 PM
Since soap and detergent action is hindered by hard water, laundry formulations usually include water softeners-called builders-designed to remove hard water ions (especially Ca2 + and Mg2 +  from the water. A common builder used in North America is sodium carbonate. Suppose that the hard water used to do laundry contains 75 ppm CaCO_3 and 55 ppm MgCO_3 (by mass).

What mass of NaCO_3 is required to remove 90.0% of these ions from 10.0 L of laundry water?
Title: Re: Using the Common Ion Effect to Remove Ions from Hard Water
Post by: Borek on March 18, 2008, 04:45:39 PM
Please read forum rules (http://www.chemicalforums.com/index.php?page=forumrules).
Title: Re: Using the Common Ion Effect to Remove Ions from Hard Water
Post by: bokwandu on March 18, 2008, 05:09:00 PM
okay so i started with converting the 75 ppm CaCO_3 and 55 ppm MgCO_3 into Mol/Liter or M since the Ksp equation uses M.  and i get .075 M for CaCO_3 and .055 M MgCO_3.

The Ksp for CaCO_3 = 4.96*10^-9 and the Ksp for MgCO_3 = 6.82*10^-6.

I begin to setup my ICE tables as

      CaCO_3        =       Ca      +     CO_3
I     .075 M                                     X
C
E

      MgCO_3      =        Mg      +     CO_3
I     .055 M                                     X
C
E

X is for the CO_3 dissolved from the NaCO_3 and err im kinda stuck onto what to do next. or am i heading already in the wrong direction?
Title: Re: Using the Common Ion Effect to Remove Ions from Hard Water
Post by: Borek on March 18, 2008, 05:38:31 PM
Quote from: bokwandu on March 18, 2008, 05:09:00 PM
i started with converting the 75 ppm CaCO_3 and 55 ppm MgCO_3 into Mol/Liter or M since the Ksp equation uses M.  and i get .075 M for CaCO_3 and .055 M MgCO_3.

Check these values - what does ppm stand for? Also note, that ppm usually means w/w (see http://www.chembuddy.com/?left=concentration&right=ppm-ppb-ppt ).

Quote
NaCO_3

Huh?

Quote
and err im kinda stuck onto what to do next. or am i heading already in the wrong direction?

No need for ICE table. Check at what concentration of CO32- you are left with 10% of Mg2+, then at waht concentration of CO32- you are left with 10% of Ca2+...
Title: Re: Using the Common Ion Effect to Remove Ions from Hard Water
Post by: bokwandu on March 19, 2008, 12:22:31 AM
My TA told me "ppm is not directly translated into molarity.  ppm means grams of 
solute/million grams of solution.  Since it's so dilute you can assume this  means
grams
of solute/million mL of solution.  Also, since we can assume  that all salts
in infinitely dilute solutions completely dissolve into their  ions, then the
moles of CaCO3 in solution equals the moles of CO3^2- in  solution from the
MgCO3, and  of the sum of the moles of CaCO3 and  MgCO3 in the solution is
determined by the total moles of CO3^2- in  the solution.  This number of moles is
the same as the  number of moles of Na2CO3 required, because the Na2CO3 
provides the CO3^2-.  I don't see the purpose of any ICE table in this 
calculation. "

I tried using his logic and i ended up getting 1.62 grams which is not correct... -_-;;;

I've also tried adding the two equations, multiplying their Ksp values,
substituting, and solving the quadratic to determine Ca2+ and Mg2+ values
and final CO3 value. But i get the wrong answer with that method too. (
10.9 grams)
Title: Re: Using the Common Ion Effect to Remove Ions from Hard Water
Post by: Borek on March 19, 2008, 04:37:10 AM
No idea what your TA meant. Have you tried appraoch I have suggested?