Question

State whether a 0.05M of solution of each of the following will be acidic , neutral or basic , explain why (use reactions) and write the structural formulas of the ions present in each solution :
a. H3PO4
b. NaH2PO4
c. Na2HPO4
d. Na3PO4
e. NH4CL
f. monosodium citrate
g. Phenol
h. NaHCO3
I. disodium succinate
j. Trisaminomethane•hydrochloride
k. methylamine.
Calculate the pH of each solution.

Answer 1

a) H₃PO₄ - acidic

H₃PO₄ = 3H⁺ + PO₄^3-

H₃PO₄ + 3NaOH = Na₃PO₄ + 3H₂O

pH = 1/2pK_a,H3PO4 - 1/lgC_H3PO4 = 1/2*2.15 - 1/2*lg(0.05) = 1.7

b) NaH₂PO₄ - both acidic and basic

NaH₂PO₄ = Na⁺ + H₂PO₄^-

NaH₂PO₄ + 2NaOH = Na₃PO₄ + 2H₂O

NaH₂PO₄ + HCl = H₃PO₄ + NaCl

pH = 1/2pK_a,H3PO4 + 1/2pK_a,H2PO4- = 1/2*2.15 + 1/2*7.21 = 4.7

c) Na₂HPO₄ - both acidic and basic

Na₂HPO₄ = 2Na⁺ + HPO₄^2-

Na₂HPO₄ + NaOH = Na₃PO₄ + H₂O

Na₂HPO₄ + 2HCl = H₃PO₄ + 2NaCl

pH = 1/2pK_a,H2PO4- + 1/2pK_a,HPO42- = 1/2*7.21 + 1/2*12.3 = 9.8

d) Na₃PO₄ - basic

Na₃PO₄ = 3Na⁺ + PO₄^3-

PO₄^3- + H₂O = HPO₄^2- + OH^-

pH = 14 - 1/2pK_b,PO43- + 1/2lgC_PO43- = 14 - 1/2*1.7 + 1/2lg(0.05) = 12.5

pK_b,PO43- = 14 - pK_a,HPO42- = 14 - 12.3 = 1.7

e) NH₄Cl - acidic

NH₄Cl = NH₄⁺ + Cl^-

NH₄⁺ + H₂O = NH₃ + H₃O⁺

pH = 1/2pK_a,NH4+ - 1/2lgC_NH4+ = 1/2*9.25 - 1/2*lg(0.05) = 5.3

pK_a,NH4+ = 14 - pK_b,NH3 = 14 - 4.75 = 9.25

f) NaC₆H₇O₇ - both acidic and basic

NaC₆H₇O₇ = Na⁺ + C₆H₇O₇^-

NaC₆H₇O₇ + 2N7aOH = Na₃C₆H₅O₇ + 2H₂O

NaC₆H₇O₇ + HCl = C₆H₈O₇ + NaCl

pH = 1/2pK_a,H2A + 1/2pK_a,HA- = 1/2*2.92 + 1/2*4.28 = 3.6

g) C₆H₅OH - acidic

C₆H₅OH = C₆H₅O^- + H⁺

C₆H₅OH + NaOH = C₆H₅ONa + H₂O

pH = 1/2pK_a,C6H5OH - 1/2lgC_C6H5OH = 1/2*10 - 1/2*lg(0.05) = 5.65

h) NaHCO₃ - both acidic and basic

NaHCO₃ = Na⁺ + HCO₃^-

NaHCO₃ + NaOH = Na₂CO₃ + H₂O

NaHCO₃ + HCl = H₂CO₃ (CO₂ + H₂O) + NaCl

pH = 1/2pK_a,H2CO3 + 1/2pK_a,HCO3- = 1/2*6.35 + 1/2*10.32 = 8.3

i) Na₂C₄H₄O₄ - basic

Na₂C₂H₂O₄ = 2Na⁺ + CH₄O₄^2-

C₂H₄O₄^2- + 2H₂O = C₂H₆O₄ + 2OH^-

pH = 14 - 1/2pK_b,C2H4O42- + 1/2lgC_C2H4O42- = 14 - 1/2*8.36 + 1/2lg(0.05) = 9.2

pK_b,C2H4O42- = 14 - pK_a,C2H5O2- = 14 - 5.64 = 8.36

j) C₄H₉O₃NH₃Cl - acidic

C₄H₉O₃NH₃Cl = C₄H₉O₃NH₃⁺ + Cl^-

C₄H₉O₃NH₃⁺ = C₄H9O₃NH₂ + H⁺

pH = 1/2pK_a,C4H9O3NH3+ - 1/2lgC_C4H9O3NH3+ = 1/2*8.075 - 1/2*lg(0.05) = 4.7

k) CH₃NH₂ - basic

pH = 14 - 1/2pK_b,CH3NH2 + 1/2lgC_CH3NH2 = 14 - 1/2*3.34 + 1/2lg(0.05) = 11.7