CHEM1210: Different salts have different solubilities in water. For some salts,
ID: 539566 • Letter: C
Question
CHEM1210:
Different salts have different solubilities in water. For some salts, the addition of H^+ or OH^- to the solution can actually increase its solubility. As a chemist for an agricultural products company, you have just developed a new herbicide, "Herbigon, " that you think has the potential to kill weeds effectively. A sparingly soluble salt, Herbigon is dissolved in 1 M acetic acid for technical reasons having to do with its production. You have determined that the solubility product K_sp of Herbigon is 9.30 times 10^-6. Although the formula of this new chemical is a trade secret, it can be revealed that the formula for Herbigon is X-acetate (XCH_3 COO, where "X" represents the top-secret cation of the salt). It is this cation that kills weeds. Since it is critical to have Herbigon dissolved (it won't kill weeds as a suspension), you are working on adjusting the pH so Herbigon will be soluble at the concentration needed to kill weeds. What pH must the solution have to yield a solution in which the concentration of X^+ is 4.00 times10^-3 M? The pK_a of acetic acid is 4.76. Express your answer numerically.Explanation / Answer
You need to maximize the dissociation of acetic acid, which will push back the amount of acetate dissociating from the mystery compound.
Ksp = [X+][C2H3O2-] = 9.3X10-6
Substituting in the desired value for [X+]
9.3X10-6 = (4.0X10-3)x
x, or acetate concentration, is . 0.002325
Now, we calculate the Ka of acetic acid:
pKa = -logKa
4.76 = -logKa
Ka = 1.738X10-5
We want the concentration of acetate to be 0.002325 , so we plug that into our acid dissociation constant expression:
Ka = [H+][C2H3O2-]/[HC2H3O2]
1.738X10-5 = x(0.002325)/1
x = 0.00748= [H+]
pH = -log[H+]
pH = -log(0.00748)
pH = 2.126