Data and calculations: The Solubility Product of Ba(O32 The calculations here ar

Question

Data and calculations: The Solubility Product of Ba(O32 The calculations here are similar to those in Experiment 23.and are outlined in the Experimental Procedure section. To use a spreadsheet, set up a table like the one below. Test Tube Number ml, 0.0350 M KIO 1.00 a 00 3.00 4.00 6 00 ml 0.0200 M Noi, 500 5.00 5do 5 5.00 Total volume in ml. 12 12 12h 12- Absorbance of solm. [lo, in soln. Processing the Data Initial moles Ba a Initial moles IO, X. 0.012 Moles IO, in equil. soln. (12 mL) 4 Moles IO, precipitated

Explanation / Answer

Moles of Ba2+ in equilibrium in 12 ml = 10 *10^5-4.8 * 10^-5

= 5.2 * 10^5

same as for all test tube

5.5 * 10 ^-5

5.3 * 10 ^-5

5.4 *10^-5

5.1 * 10^-5

Molarity of the Ba^2+ in equiibrium = number of moles/ volume of solotion

5.2 *10-5/0.012L

=433.3 * 10-5 mOLES/L

same as all

458.3 *10-5

441 * 10^-5

450 *10-5

425*10-5

Ksp for Ba(IO3)2

Ba(IO3)2 ---------> Ba^2+ + 2IO3^-1

s (2s)^2

Ksp = 4s3

where s stands for the molarity of the Ba and IO3-

1. ) 4*( 433.3* 10 -5)3

= 32.5 *10-8

2) 38.48 *10-8

3) 34.30*10-8

4) 36.45 *10-8

5) 30.7 *10-8

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