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Report (any type)/Brief report
22/09/2021
##2.9xx10^(-2) M##
Let’s start by writing the chemical reaction for the dissociation of silver phosphate:
##Ag_3PO_4(s) rightleftharpoons 3Ag^(+)(aq) +PO_4^(-3)(aq)##
Now, set the value equal to the products (you don’t care about the reactants because it’s a solid). Ag is raised to the 3rd power because the coefficient is 3.
##Ksp=1.8xx10^(-5) M = [Ag^(+)]^(3) (aq) +[PO_4^(3-)] (aq)##
Replace each reactant with the letter x because that’s what you’re trying to find. Since the coefficient in front of silver is 3, a 3 must be placed in front of the x and it must be raised to the 3rd power.
##1.8xx10^(-5) M = (3X)^(3) xx (X)## (always multiply when finding the molar solubility).
Now take ##3^(3)## which is 27 and divide the Ksp by 27, so you can get all of the X’s by themselves.
##(1.8xx10^(-5) M)/27 ## = ##6.67xx10^(-7)M##
Now you’re left with ##X^(3) xx X##, so multiply the X’s to get ##X^(4)##. Take the fourth root of the ##6.67xx10^(-7)M## to obtain the value of x.
(##6.67xx10^(-7)M)^(1/4) = 2.9xx10^(-2) M##
The value of x that we just obtained is our molar solubility.
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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