Question

If 82.5 mL of 0.723 M `H_2SO_4` titrates completely with 34.0 mL of `Al(OH)_3` solution, what is the molarity of the `Al(OH)_3` solution?

Answer 1

`1.17 (mol)/L`

Explanation:

First, let's write the equation for this titration reaction:

`3H_2SO_4 (aq) + 2Al(OH)_3 (aq) rarr Al_2(SO_4)_3 (aq) + 6H_2O (l)`

The solubility of `Al(OH)_3` is relatively low, but to simplify this we'll still use `(aq)` in the equation.

For any titration problem like this, where you try and find the concentration of one substance when you know the other concentration and volumes of both, the next step is to convert the standard solution (the one with known volume and concentration) to moles:

`molH_2SO_4 = (0.0825L)(0.723 (mol)/L) = 0.0596 mol H_2SO_4`

Now we'll use the stoichiometrically equivalent values to find the moles of the substance with unknown concentration, `Al(OH)_3`:

`0.0596 mol H_2SO_4((2 mol Al(OH)_3)/(3 mol H_2SO_4)) = 0.0397 mol Al(OH)_3`

Finally, we'll use this value and its known volume to find its molar concentration:

`M = (0.0397 mol Al(OH)_3)/(0.0340 L) = 1.17 (mol)/L`