The lab technician Anna Lytic adds 2.20 mol KOH to 1.00 L of 0.5 M Al(NO₃)₃.What is the concentration of aluminum ions after the aluminum nitrate has reacted with the potassium hydroxide? K_f = 3.0 $\times$ 10³³ for Al(OH)₄^- A)1.8 $\times$ 10^-7 M B)9.l $\times$ 10^-18 M C)1.0 $\times$ 10^-31 M D)3.3 $\times$ 10^-34 M E)7.l $\times$ 10^-36 M

Question

Quizplus · Accepted Answer

First, we need to write out the balanced chemical equation for the reaction between Al(NO3)3 and KOH:

Al(NO3)3 + 3KOH → Al(OH)3 + 3KNO3

We can see that for every 1 mole of Al(NO3)3, we will produce 1 mole of Al(OH)3. Using the stoichiometry of the reaction, we can calculate the number of moles of Al(OH)3 produced:

2.20 mol KOH x (1 mol Al(NO3)3 / 3 mol KOH) x (1 mol Al(OH)3 / 1 mol Al(NO3)3) = 0.733 mol Al(OH)3

Now we use the formula given in the problem to calculate the concentration of aluminum ions:

[Al3+] = 3.0 x 10^-10 x [Al(OH)3]^3

[Al3+] = 3.0 x 10^-10 x (0.733 M)^3

[Al3+] = 1.0 x 10^-31 M

Therefore, the concentration of aluminum ions is 1.0 x 10^-31 M, which corresponds to choice C.

The Lab Technician Anna Lytic Adds 2 ×\times× 1033 for Al(OH)4- A)1

The Lab Technician Anna Lytic Adds 2 $\times$ 10³³ for Al(OH)₄^-
A)1