Asked by Student12
Calculate the mass of solid AgNo3 that can be added to 2.0 L of 0.10M K2CrO4 solution in order to just start precipitation?
Here is what I have done so far:
2AgNO3(aq) + K2CrO4(aq) ==> Ag2CrO4(s) + 2KNO3(aq)
Ag2CrO4(s) ===> 2Ag^+ CrO4^^2-
Ksp = [Ag^+]^2[CrO4^2-]
Ksp = (2x)^2(0.1)
Please help me in what do next and show me how to get to answer.
Here is what I have done so far:
2AgNO3(aq) + K2CrO4(aq) ==> Ag2CrO4(s) + 2KNO3(aq)
Ag2CrO4(s) ===> 2Ag^+ CrO4^^2-
Ksp = [Ag^+]^2[CrO4^2-]
Ksp = (2x)^2(0.1)
Please help me in what do next and show me how to get to answer.
Answers
Answered by
DrBob222
Ksp = (2x)^2(0.1)
You can look up the Ksp for Ag2CrO4. It's approximately 9E-12 but you should use the value in your book's table of Ksp values.
9E-12 = 4x^2(0.1)
9E-12/0.1 = 4x^2
9E-11 = 4x^2
9E-12/4 = 2.25E-11 = x^2
x = sqrt 2.25E-11 = 4.47E-6 M
In 2L that is 4.47E-6 x 2 = ?
Then grams AgNO3 = mols rom above x molar mass AgNO3.
You can look up the Ksp for Ag2CrO4. It's approximately 9E-12 but you should use the value in your book's table of Ksp values.
9E-12 = 4x^2(0.1)
9E-12/0.1 = 4x^2
9E-11 = 4x^2
9E-12/4 = 2.25E-11 = x^2
x = sqrt 2.25E-11 = 4.47E-6 M
In 2L that is 4.47E-6 x 2 = ?
Then grams AgNO3 = mols rom above x molar mass AgNO3.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.