Answer to Question #113903 in General Chemistry for Layali Hassouneh

Question

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? Kf = 3.0 × 1033 for Al(OH)4-

Answer 1

As moles of KOH = moles of OH- = 2.20 mol

and , moles of Al(NO3)3 = moles of Al3+ = 0.5 mol

Now , Balanced reaction between Al3+ and OH- is

Al3+ + 4OH- -----> Al(OH)4-

Now , Final moles of OH- after reaction

= Initial moles of OH- - = 20 - 2.00 = 0.20 mol

As volume of solution = 1.0 L

So , Final conc. of OH- = [OH-] = 0.20 / 1 = 0.20 M

Also , Final moles of Al(OH)4- = Initial moles of Al3+ = 0.5 mol

So , Final Conc. of Al(OH4)- = 0.5 / 1 = 0.5 M

As Kf = [Al(OH)4- ] / ( [Al3+] [OH-]4 )

So , [Al3+] = [Al(OH)4-] / ( kf × [OH-]4 )= 1.0*10-31