A solution is made by adding potassium phosphate (K_3PO_4 (MW = 134 g/mol; EW =

Question

A solution is made by adding potassium phosphate (K_3PO_4 (MW = 134 g/mol; EW = 44.67 g/eq)) to pure water to give a concentration of 10^-1 M. Assume the K_3PO_4 completely dissolves and dissociates upon addition to water (it is a salt). (a) Write all chemical equilibrium reactions for this solution. (b) Using the law of mass action, write all chemical equilibrium expressions (equations) for this solution. (c) Write the mass balance equation(s) for this solution. (d) Write the charge balance for this solution. (e) Estimate the pH of the solution.

Explanation / Answer

Answer 7 (a) : Chemical equilibrium reactions that take place upon dissolution of K3PO4 in water are

1 ) K3PO4 <=======> K+ + K2PO4^-

2 )    K2PO4^- <=======> K+ + KPO4^2-

3) KPO4^2- <=======> K+ + PO4^3-

Answer 7 (b) Chemical equilibrium expression for reaction (1)

K = [ K+] [K2PO4^-] / [K3PO4]

  Chemical equilibrium expression for reaction (2)

K = [ K+] [KPO4^2-] / [K2PO4^-]

  Chemical equilibrium expression for reaction (3)

K = [ K+] [PO4^3-] / [KPO4^2-]

Answer 7 (c) Since solution made is 0.1 M that means 0.1 moles of K3PO4 are added per litre of solution.

K3PO4 <=======> 3K + PO4

0.1 mole 0.3 moles 0.1 mole

0.1 mol x 212g/mol 0.3 mol x 39.1 g/mol 0.1 mol x 95 g/mol

21.2 g 11.73 g 9.5 g

Mass of reactant = 21.2 g

Mass of products = 11.73 g + 9.1 g = 21.2 g

Note : Molecular weight of K3PO4 is 212 g / mol.

Answer 7 (d) : K3PO4 <======> 3K^+ + PO4^3-

Answer 7 (e) : PO4^3- + H2O =======> HPO4^2- + OH^-

I (M) 0.1 0 0

C (M) -x +x +x

E (M) 0.1 - x x x

Kb = x^2 / 0.1 - x = 1.45 x 10^-12 (x in the denominator can be ignored)

x = 0.380 x 10^-6 = 3.80 x 10^-6

pOH = - log (3.80 x 10^-6) = 5.42

pH = 14 - pOH = 8.57

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.