Question

The colorless gas phosphorus trifluoride reacts slowly with water to give a mixture of phosphorus acid and hydrofluoric acid . (a) Write a balanced chemical equation for this reaction. + + (b) Determine the concentration (in moles per liter) of each of the acids that result from the complete reaction of 5.71×10-2 moles of phosphorus trifluoride with enough water to give a solution volume of 711 mL. phosphorus acid : M hydrofluoric acid : M

Answer 1

Answer:

The balanced chemical reaction is given as:

$PF_3(g)+3H_2O(l)\rightarrow H_3PO_3(aq)+3HF(aq)$

b) Concentration of phosphorus acid is 0.0803 M.

Concentration of hydrofluoric acid is 0.241 M.

Explanation:

a) The balanced chemical reaction is given as:

$PF_3(g)+3H_2O(l)\rightarrow H_3PO_3(aq)+3HF(aq)$

b)

Moles of phosphorus trifluoride gas = $5.71\times 10^{-2} mol$

$PF_3(g)+3H_2O(l)\rightarrow H_3PO_3(aq)+3HF(aq)$

According to reaction, 1 mole of phosphorus trifluoride gives 1 mole of phosphorus acid, then $5.71\times 10^{-2} mol$ of phosphorus trifluoride will give:

$\frac{1}{1}\times 5.71\times 10^{-2} mol= 5.71\times 10^{-2} mol$ of phosphorus acid

Volume of the solution = V = 711 mL = 0.711 L ( 1 mL = 0.001 L)

$[concentration]=\frac{Moles}{Volume(L)}$

Concentration of phosphorus acid :

$[H_3PO_3]=\frac{5.71\times 10^{-2} mol}{0.711 L}=0.0803 M$

Concentration of phosphorus acid is 0.0803 M.

According to reaction, 1 mole of phosphorus trifluoride gives 3 mole of hydrofluoric acid, then $5.71\times 10^{-2} mol$ of phosphorus trifluoride will give:

$\frac{3}{1}\times 5.71\times 10^{-2} mol= 0.1713 mol$ of hydrofluoric acid

Volume of the solution = V = 711 mL = 0.711 L ( 1 mL = 0.001 L)

$[concentration]=\frac{Moles}{Volume(L)}$

Concentration of hydrofluoric acid :

$[HF]=\frac{0.1713 mol}{0.711 L}=0.241 M$

Concentration of hydrofluoric acid is 0.241 M.