Answer to Question #122142 in Chemistry for Somila

Question

Describe the preparation of the following solution (show all calculations): 5.0 L of 6.000 M H3PO4 from a concentrated solution that is 85.1%(m/m) and has a specific gravity of 1.69

Answer 1

In order to prepare $V$= 5 L of $c$ = 6.000 M H₃PO₄ solution, one needs the following quantity $n$ of H₃PO₄:

$n(H_3PO_4) = cV=5·6.000 = 30$ mol.

The mass of this quantity of H₃PO₄ can be calculated using the molar mass $M$ of H₃PO₄ (98.00 g/mol):

$m = nM = 30·98.00 =2940$ g.

Thus, the mass of the concentrated solution needed contains 2940 g of H₃PO₄:

$m_s = \frac{2940}{0.851} = 3454.8$ g.

The specific gravity of the concentrated solution is 1.69 g/mL. Therefore, 3454.8 g of the concentrated solution corresponds to the following volume:

$V_s = \frac{m}{d} = \frac{3454.8}{1.69}=2044$ mL.

Then, to 2044 mL of the concentrated solution one needs to add the following volume of water:

$V_w = 5000-2044 = 2956$ mL.

Finally, in order to prepare 5 L of 6.000 M of H₃PO₄ from a concentrated solution

that is 85.1%(m/m) and has a specific gravity of 1.69 g/mL, one needs to take 2044 mL of the concentrated solution and add 2956 mL of water.