Question

Back Titration (determining moles OH- and amount of Al(OH)3/Mg(OH)2)?

A 500mg tablet containing an antacid mixture of aluminum hydroxide, magnesium hydroxide, and inert material was dissolved in 50.0 mL of 0.500 M HCl solution. The resulting solution, which was still acidic, required 26.5 mL of 0.377 M NaOH for neutralization.

a) How many moles of OH- were in the tablet?
So far, I determined the moles of HCl that reacted by taking the difference between the moles of HCl titrated and the total moles of HCl (both found with mol = Molarity x Volume) to get 0.015 mol.

0.025 mol - 0.00999 mol = 0.015 mol HCl reacted

Apart from this, I am lost. I tried coming up with some equations for the mixture of aluminum hydroxide and magnesium to have hydronium and hydroxide ions present on the product side but nothing seemed to balance or work.

b) If the tablet contained equal moles of aluminium hydroxide and magnesium hydroxide, how many grams of each were in the tablet?

Answer 1

(a) The tablet contained 15.0 mmol `"OH"^"-"`.
(b) The tablet contained 234 mg `"Al(OH)"_3` and 175 mg `"Mg(OH)"_2`

Explanation:

(a) Moles of `"OH"^"-"`

`"Moles of HCl" = 50.00 color(red)(cancel(color(black)("mL HCl"))) × "0.500 mmol HCl"/(1 color(red)(cancel(color(black)("mL HCl")))) = "25.00 mmol HCl"`

`"Moles of NaOH" = 26.5 color(red)(cancel(color(black)("mL NaOH"))) × "0.377 mmol NaOH"/(1 color(red)(cancel(color(black)("mL NaOH")))) = "9.991 mmol NaOH"`

`"Excess moles of HCl" = 9.991 color(red)(cancel(color(black)("mmol NaOH"))) × "1 mmol HCl"/(1 color(red)(cancel(color(black)("mmol NaOH")))) = "9.991 mmol HCl"`

`"Moles HCl reacted = Total moles - excess moles"`

`" = 25.00 mmol - 9.991 mmol = 15.01 mmol"`

`"Moles of OH"^"-" = 15.01 color(red)(cancel(color(black)("mmol HCl"))) × ("1 mmol OH"^"-")/(1 color(red)(cancel(color(black)("mmol HCl")))) = "15.0 mmol OH"^"-"`

The tablet contained 15.0 mmol `"OH"^"-"`.

(b) Masses of `"Al(OH)"_3` and `"Mg(OH)"_2`

The equations are:

`"Al(OH)"_3 → "Al"^"3+" + 3"OH"^"-"`
`"Mg(OH)"_2 → "Mg"^"2+" + 2"OH"^"-"`

If there are equal moles of `"Al(OH)"_3` and `"Mg(OH)"_2`, then three-fifths of the hydroxide ions (9.01 mmol) come from aluminium hydroxide, and two fifths
(6.00 mmol) come from magnesium hydroxide.

`"Moles of Al(OH)"_3 = 9.01 color(red)(cancel(color(black)("mmol OH"^"-"))) × ("1 mmol Al(OH)"_3)/(3 color(red)(cancel(color(black)("mmol OH"^"-")))) = "3.00 mmol Al(OH)"_3`

`"Mass of Al(OH)"_3 = 3.00 color(red)(cancel(color(black)("mmol Al(OH)"_3))) × "78.00 mg Al(OH)"_3/(1 color(red)(cancel(color(black)("mmol Al(OH)"_3)))) = "234 mg Al(OH)"_3`

`"Moles of Mg(OH)"_2 = "Moles of Al(OH)"_3 = "3.00 mmol"`

`"Mass of Mg(OH)"_2 = 3.00 color(red)(cancel(color(black)("mmol Mg(OH)"_2))) × ("58.32 mg Mg(OH)"_2)/(1 color(red)(cancel(color(black)("mmol Mg(OH)"_2)))) = "175 mg Mg(OH)"_2`

The tablet contained 234 mg `"Al(OH)"_3` and 175 mg `"Mg(OH)"_2`.