A 0.2550-g sample of impure Ca(OH)2 is dissolved in enough water to make 62.00 mL of solution. 20.00 mL of the resulting solution is then titrated with 0.2333-M HCl. What is the percent purity of the calcium hydroxide if the titration requires 7.52 mL of the acid to reach the endpoint?

here's what i'm doing. maybe you can point out what's wrong. Determine millimoles H30^1+ (20.0 mL x .2333 = 4.666 millimoles H30^1+) Determine millimoles OH^1- (its a 1-1 ratio so its still 4.666). Convert mmol OH^1- to mass of Ca(OH)2 in 7.52 mL soln (4.666*74.08 mg=345.66 mg Ca(OH)2). Then take into account dilution to find mass in 62 mL solution (62mL x (345.66/7.52)=2.85 grams) then its just 2.85/.2550 x 100 to find percent, but i'm getting a rediculously high % that is wrong.

1/31/2007 8:07:49 PM

I believe that you have the dilution factors wrong. Try this:

Step 1: all Ca(OH)2 goes into the 62.00 mL soln.
Step 2: An 20.00 aliquot of that solution is removed and titrated.

Now, you have 345.7 mg of Ca(OH)2 determined to have been in the aliquot (you've used the 7.52 mL of acid, so that's done with).

So, how do you determine how many mg of Ca(OH)2 are in 62.00 mL of the same concentration of solution? Recall that taking an aliquot (sample) does not change the concentration.

Hope this helps.

[Edited on January 31, 2007 at 8:37 PM. Reason : subject/verb agreement]

1/31/2007 8:28:30 PM

I was so proud of myself when I figured out how to do titration problems in this class.

1/31/2007 11:01:23 PM

hehe

tit

1/31/2007 11:12:10 PM

^haha

2/1/2007 7:59:04 PM