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molar solubility
250.0 mL of a solution which is 1.00 M in an unknown acid is mixed with 250.0 mL of a solution which is 1.00 M in its conjugate base. 50.00g of solid AgC2H3O2 is added to this solution. After thorough mixing the solution is assumed to be saturated. To determine the amount which remains undissolved, the solution is filtered and the residue is dried and weighed; the mass of the undissolved AgC2H3O2 is 26.102 g.Calculate the molar solubility of AgC2H3O2 in this buffered solution.
Calculate the conditional Ksp value for AgC2H3O2 in this buffered solution.
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- By chem116 |
- Category: chemistry |
- Grade level: college and above
- Aug/04/2008 |
- Answers to this question: 1
|
Answer by: scott8148
First Answer
23.90g (50.00-26.102 ) of silver acetate are dissolved in 500.0ml (250.0+250.0) of solution
the molar mass of silver acetate is 166.91g
__ so .1432 moles (23.90/166.91) are in .5000L
the molar solubility is .1432/.5000 or .2864M
the buffering action of the solution results in the acetate ions being converted to acetic acid which has a dissociation constant (Ka) of 1.738x10^-5
let x="acetate concentration", so Ka=x^2/[.2864-x] __ .2684Ka-Ka(x)=x^2 __ 0=x^2+Ka(x)-.2684Ka
using quadratic formula __ x=.002993
so Ksp for silver acetate is .2864*.002993 or .0008572
the molar mass of silver acetate is 166.91g
__ so .1432 moles (23.90/166.91) are in .5000L
the molar solubility is .1432/.5000 or .2864M
the buffering action of the solution results in the acetate ions being converted to acetic acid which has a dissociation constant (Ka) of 1.738x10^-5
let x="acetate concentration", so Ka=x^2/[.2864-x] __ .2684Ka-Ka(x)=x^2 __ 0=x^2+Ka(x)-.2684Ka
using quadratic formula __ x=.002993
so Ksp for silver acetate is .2864*.002993 or .0008572
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- Aug/04/2008 |
- Answers by scott8148: 170 |
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