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May 5, 2016

# Homework Help: chemistry

Posted by bria on Wednesday, September 13, 2006 at 2:21pm.

If a solution of 0.10 M silver nitrate, AgNO3, is slowly added to a solution containing 0.010 M potassium iodate, KIO3, and 0.20 M sodium bromate, NaBrO3, what precipitate will form first. Will this precipitate form nearly completely (99%) before the second precipitate starts to form?
Qsp of AgIO3 is 3.6 * 10^-8 and the Qsp of AgBrO3 is 5.01 * 10^-5

I think you are using Qsp incorrectly. I believe the values you list are Ksp values and not Qsp.
For the first part, write the Ksp expression for AgIO3 and for AgBrO3. Using 0.01 for (IO3^-) and 0.2 for (BrO3^-), calculate (Ag^+) necessary to cause precipitation of each salt. Since AgNO3 is being added slowly, the first ppt will be that requiring the smaller (Ag^+). Let's digest this one before we start on the 2nd part. I hope this helps

For the second part, technically, I don't think it can be done exactly because there are no volumes given for KIO3 and KBrO3. However, since the molarities of KIO3 and KBrO3 are so different, I suspect that the author assumed dilution would not play a factor and you are to make this comparison based on molarity at the beginning and at the end. If I have assumed correctly, then proceed by taking the ratio of the Ksp values.
[(Ag^+)(IO3^-)/(Ag^+)(BrO3^-)]=[Ksp AgNO3/Ksp AgBrO3]= 3.6E-8/5.01E-5.

Note that (Ag^+) cancels.
Solve for (IO3^-) = ?? x (BrO3^-)
At the moment just before the first molecule of AgBrO3 ppts, (BrO3^-) = 0.2 (assuming no dilution effects in pptg the AgIO3 at the beginning) and this will give you the molarity of the (IO3^-) at that point before any AgBrO3 has pptd.

Then compare that to the 0.01M it was at the beginning. Doing this results in about 98.5% (approximately--you will need to do it more exactly) pptn (or about 1.5% not pptd). Check my thinking. Check my arithmetic. In a separate process, I assumed some volumes, such as 50 mL KIO3 and 50 mL KBrO3 and worked through the dilution part and came up with only minor differences; i.e., only about 0.01% difference, and that makes me think the author assumed no dilution effects. I hope this helps.

Thank you so much for you help.

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