Question

Calculate the ionic strength of a 48.0 wt% phosphoric acid, if the phosphoric acid has a density of 2.4 g/mL.

Answer 1

Density is 2.4 g/mL

1000 mL (1 L ) of solution contains $2.4 \times 1000 = 2400$ g of solution.

The weight percent of solution is 48.0 wt%

2400 g of solution contains $2400 \times \dfrac {48.0}{100} = 1152$ g of phosphoric acid.

The molecular weight of phosphoric acid is 98 g/mol.

The number of moles of phosphoric acid $= \dfrac { 1152 }{98}=11.8$ moles.

Since total volume is 1 L, the number of moles is equal to molar concentration.

$[H_3PO_4]=[PO_4^{3-}]=11.8 \: M \\ \: [H^+] = 3 \times 11.8 \: M = 35.3 \: M$

The ionic strength

$I = \dfrac {1} {2} \sum_{1}^{n} c_iz_i^2 = \dfrac {1} {2} \left [ \left ( 35.3 \times 1^2 \right ) +11.8 \times 3^2\right ] \\ \: I = \dfrac {1}{2} \left [ 35.3+106.2 \right ]=70.8 \: M$