# 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?

May 16, 2017

$1.17 \frac{m o l}{L}$

#### Explanation:

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

$3 {H}_{2} S {O}_{4} \left(a q\right) + 2 A l {\left(O H\right)}_{3} \left(a q\right) \rightarrow A {l}_{2} {\left(S {O}_{4}\right)}_{3} \left(a q\right) + 6 {H}_{2} O \left(l\right)$

The solubility of $A l {\left(O H\right)}_{3}$ is relatively low, but to simplify this we'll still use $\left(a q\right)$ 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:

$m o l {H}_{2} S {O}_{4} = \left(0.0825 L\right) \left(0.723 \frac{m o l}{L}\right) = 0.0596 m o l {H}_{2} S {O}_{4}$

Now we'll use the stoichiometrically equivalent values to find the moles of the substance with unknown concentration, $A l {\left(O H\right)}_{3}$:

$0.0596 m o l {H}_{2} S {O}_{4} \left(\frac{2 m o l A l {\left(O H\right)}_{3}}{3 m o l {H}_{2} S {O}_{4}}\right) = 0.0397 m o l A l {\left(O H\right)}_{3}$

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

$M = \frac{0.0397 m o l A l {\left(O H\right)}_{3}}{0.0340 L} = 1.17 \frac{m o l}{L}$