Question

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 Ksp of Herbigon is 9.40×10−6. Although the formula of this new chemical is a trade secret, it can be revealed that the formula for Herbigon is X-acetate (XCH3COO, 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 5.00×10−3 M ? The pKa of acetic acid is 4.76.

Answer 1

Answer:

pH = 2.03

Explanation:

The pH can be calculated using the following equation:

$pH = -log [H_{3}O^{+}]$  (1)

The concentration of H₃O⁺ is calculated using the dissociation constant of the next reaction:

CH₃COOH + H₂O ⇄  CH₃COO⁻ + H₃O⁺    

   1.00 M    

$K_{a} = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$

Solving the above equation for H₃O⁺, we have:    

$[H_{3}O^{+}] = \frac{Ka*[CH_{3}COOH]}{[CH_{3}COO^{-}]}$    (2)    

The dissociation constant is equal to:    

$pKa = -log(Ka) \rightarrow Ka = 10^{-pKa} = 10^{-4.76} = 1.74 \cdot 10^{-5}$    

Now, by solving the equation of the solubility product for Herbigon, we can find [CH₃COO⁻]:

CH₃COOX  ⇄  CH₃COO⁻ +  X⁺  

                                             5.00x10⁻³ M

$K_{sp} = [CH_{3}COO^{-}][X^{+}]$

$[CH_{3}COO^{-}] = \frac{K_{sp}}{[X^{+}]} = \frac{9.40 \cdot 10^{-6}}{5.00 \cdot 10^{-3}} = 1.88 \cdot 10^{-3} M$

By entering the values of [CH₃COO⁻] and Ka, into equation (2) we can calculate [H₃O⁺]:

$[H_{3}O^{+}] = \frac{1.74 \cdot 10^{-5}*[1.00]}{[1.88 \cdot 10^{-3}]} = 9.26 \cdot 10^{-3} M$

Hence, the pH is:

$pH = -log [H_{3}O^{+}] = -log [9.26 \cdot 10^{-3}] = 2.03$

Therefore, the pH must be 2.03 to yield a solution in which the concentration of X⁺ is 5.00x10⁻³M.

I hope it helps you!