Question

The Ksp can be used to find the concentration of hydroxide ions, and thus determine the pH given a concentration of of aluminum. At what pH will precipitation of Al(OH)3 begin if 6.70 lb of aluminum sulfate, Al2(SO4)3, is added to 2050 gallons of water (with a negligible change in volume)?

Answer 1

Answer:

3.9.

Explanation:

(Molar mass of 342.15 g/mol).

The first thing to do in this question is to coverts our units. We are given 6.70 lb = 453.592 × 6.70 lb = 3039.0664 g of aluminum sulfate.

2050 gallons of water = 2050 × 3.785 = 7759.25 Litres.

After this we will now look for the Concentration aluminum sulfate using the formula below:

Concentration = number of moles ÷ volume. ------------------------------------(1).

Recall; Number of moles = 3039.0664/ 342.15 = 8.8823 moles. The volume = 2050 gallons = 7759.25 Litres.

Therefore, slotting in the values of of number of moles and volume into the equation (1) above, we have;

Concentration= 8.8823 moles/ 7759.25 Litres. = 0.001145 M.

The next equation we are going to make use of today is that of the solubilty product that is;

Ksp = (Al^3+) (OH^-)^3.

So, we have the ksp = 1.3 × 10^-33 and the value of (Al^3+)= 0.001145 M.

Hence, making (OH^-) the subject of the formula;

(OH^-) = ( 1.3 × 10^-33 / 0.001145)^1/3.

(OH^-) = 8 × 10^-11 M.

Hence, pOH = - log (OH^-)

pOH = - log (8 × 10^-11).

pOH = 10.1

Therefore, the values of pH is; 1

pH + pOH = 14.

pH = 14 - 10.1 = 3.9.