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#### inquisitive

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« on: November 21, 2007, 07:43:03 PM »
I'm realy bothered at this problem in our book:

A volume of 75 ml of 0.060 M NaF is mixed with 25 ml of 0.15 M Sr(NO3)2. Calculate the concentrations in the final solution of NO3-, Na+, Sr2+, and F-. ( Ksp for SrF2 = 2.0 x 10-10.)

When I solved it for the first time and compared my answer in the back of the book,
I only got the concentrations for the Na+ and NO3- correct but wrong in the Sr2+ and F-. I solved the problem many times and still I can't arrive at the correct concentrations of Sr2+ and F-. I can't really figure out what the Ksp of SrF2 has something to do with this problem. Can someone show me how to solve this problem? (Concentrations of the species should be Na+ = 0.045, NO3- = 0.076, Sr2+ = 0.016, and F- = 1.1 x 10-4.

#### LQ43

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« Reply #1 on: November 21, 2007, 10:56:04 PM »
can you show us how you tried to solve it? (with balanced equation?) we can look at your steps and help to find where to put you on the right track.

Since SrF2 is a slightly soluble salt, some of it actually dissolves and is in solution. That is where the Ksp comes in.  A dissociation equation is needed to go along with the Ksp equation. There might be a section of solubility and equilibrium in your textbook.

#### AWK

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« Reply #2 on: November 22, 2007, 01:21:51 AM »
This is a problem with common ion effect (an excess of strontium cation)
NO3- should be 0.075,
Sr2+ should be 0.015
if you neglect ionic strenght.
AWK

#### inquisitive

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« Reply #3 on: November 22, 2007, 04:48:06 AM »
can you show us how you tried to solve it? (with balanced equation?) we can look at your steps and help to find where to put you on the right track.

Since SrF2 is a slightly soluble salt, some of it actually dissolves and is in solution. That is where the Ksp comes in.  A dissociation equation is needed to go along with the Ksp equation. There might be a section of solubility and equilibrium in your textbook.
No balanced equation was given in the problem so formulated it myself. I got this equation:
2NaF + Sr(NO3)2 ---> 2Na(NO3)2 + SrF2
These is what I did to find the sodium ion concentration:
The number of moles of Na+ present in the original 75 mL solution is

75 mL   x  1 L  x  0.06 mol NaF  x  1 mol Na
1000 mL        1 L           1 mol NaF

= 4.5 x 10-3 Na+.
The total volume after combining the two solutions is 100 mL. The concentration of Na+ in the 100 mL volume is

[Na+] = 4.5 x 10-3 mol 1000 mL
100 mL           1 L
[Na+] = 0.045.
Using the same logic I got the the concentration of NO3-
= 0.076. However, if I will further apply this to the Sr2+ and F-, I can't arrive at the correct answer.

#### AWK

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« Reply #4 on: November 22, 2007, 05:04:00 AM »
[NO3-]= 2 x 25 x 0.15 /100 = 0.075 !
AWK

#### Borek ##### Re: a precipitation reaction problem
« Reply #5 on: November 22, 2007, 06:49:26 AM »
You can't calculate Sr2+ and F- concentrations the same way, as they were removed from the solution by precipitation. Start with a limiting reagent approach to find out what is left in the solution.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### LQ43

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« Reply #6 on: November 22, 2007, 12:23:13 PM »
just a bit of organization....

1. Calculate INITIAL concentrations of Na+ and NO3-

No balanced equation was given in the problem so formulated it myself. I got this equation:
2NaF + Sr(NO3)2 ---> 2Na(NO3)2 + SrF2
These is what I did to find the sodium ion concentration:
The number of moles of Na+ present in the original 75 mL solution is

75 mL   x  1 L  x  0.06 mol NaF  x  1 mol Na
1000 mL        1 L           1 mol NaF

= 4.5 x 10-3 Na+.
The total volume after combining the two solutions is 100 mL. The concentration of Na+ in the 100 mL volume is

[Na+] = 4.5 x 10-3 mol 1000 mL
100 mL           1 L
[Na+] = 0.045.
good

[NO3-]= 2 x 25 x 0.15 /100 = 0.075 !

2.  Calculate INITIAL concentrations of Sr2+ and F- like above BUT then...

Start with a limiting reagent approach to find out what is left in the solution.

3.  then

Since SrF2 is a slightly soluble salt, some of it actually dissolves and is in solution. That is where the Ksp comes in.  A dissociation equation is needed to go along with the Ksp equation. There might be a section of solubility and equilibrium in your textbook.
This is a problem with common ion effect (an excess of strontium cation)

quite a bit of work here...

(Concentrations of the species should be Na+ = 0.045, F- = 1.1 x 10-4.
NO3- should be 0.075,
Sr2+ should be 0.015
if you neglect ionic strenght.

#### Sev

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« Reply #7 on: November 22, 2007, 06:42:42 PM »
Quote
2NaF + Sr(NO3)2 ---> 2Na(NO3)2 + SrF2

JTLYK this eqn is wrong.