Asked by joe
what is the solubility of silver carbonate in water in 25 degrees celcius if Ksp=8.4X10-12? i don't know what to do
Write the equation.
Ag2CO3(s) ==> 2Ag^+ + CO3^=
Ksp = (Ag^+2)(CO3^=)
Let x = solubility of Ag2CO3. At equilibrium,(Ag^+) = 2x; (CO3^=) = x. Plug those into the Ksp expression and solve for x. The answer comes out in mols/L. If you want grams, then M x molar mass Ag2CO3 = grams Ag2CO3.
Post your work if you get stuck.
Write the equation.
Ag2CO3(s) ==> 2Ag^+ + CO3^=
Ksp = (Ag^+2)(CO3^=)
Let x = solubility of Ag2CO3. At equilibrium,(Ag^+) = 2x; (CO3^=) = x. Plug those into the Ksp expression and solve for x. The answer comes out in mols/L. If you want grams, then M x molar mass Ag2CO3 = grams Ag2CO3.
Post your work if you get stuck.
Answers
Answered by
Annatolia
Ag2CO3= 2Ag^1+CO3^-2
Ksp=[Ag^+1]^2[CO3^-2]
. =(2x)^2(x)
8.4*10^-12=4x^3
8.4*10^-12/4= 4x^3/4
Cube root of 2.1*10^-12 =x
0.0001280
1.28*10^-4M=x
Ksp=[Ag^+1]^2[CO3^-2]
. =(2x)^2(x)
8.4*10^-12=4x^3
8.4*10^-12/4= 4x^3/4
Cube root of 2.1*10^-12 =x
0.0001280
1.28*10^-4M=x
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