A saturated solution of AgCl was found to have a concentration of 1.30 x 10-5 mol dm-3. The solubility product of AgCl therefore is?
1.30 x 10-5 mol2 dm-6
1.30 x 10-7 mol2 dm-6
1.69 x 10-10 mol2 dm-6
2.60 x 10-12 mol2 dm-6
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Ksp(solubility product)=?
the salt dissociates into
Agcl →Ag+ Cl-
the conc ratio = 1:1
we know that
Ksp = [Ag+] [Cl-]
subst Ag+, Cl- for x
Ksp = (x)(x)
Ksp = x².
but x being the conc of the ions involved. was given, =1.30x10^-5
Ksp = (1.30x10^-5)²
Ksp = 1.68x10^-10
please this answer is wrong.solubility can regarded as concentration thus can be expressas same unit as conc.that is one latent point in solution and solubilit study and calculation for jamb student confirm this at page 210 of outline chemistry 2014 edition by ojiodu C.C.so the answer is A.
It's not really that difficult to find using this method
To find the solubility product constant of Agcl:
Ksp = [Ag+] × [Cl-]
Since AgCl dissociates into equal concentrations of Ag+ and Cl-:
[Ag+] = [Cl-] = 1.30 × 10^(-5) mole/dm³
Ksp = [Ag+] × [Cl-]
= (1.30 × 10^(-5)) × (1.30 × 10^(-5))
= 1.69 × 10^(-10)
So, the solubility product (Ksp) of AgCl is:
1.69 × 10^(-10)
Note: ksp stands for solubility product constant
To get your solubility product
KsP= 1.30 * 10^-5 * 1.30 * 10^-5=. 1.69*10^-10
So answer is = 1.69*10^-10. C
