Answer to Question #99502 in General Chemistry for Boye Zephaniah

Question #99502
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.
1
Expert's answer
2019-12-05T08:19:06-0500

a) H3PO4 - acidic

H3PO4 = 3H+ + PO43-

H3PO4 + 3NaOH = Na3PO4 + 3H2O

pH = 1/2pKa,H3PO4 - 1/lgCH3PO4 = 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 = 1/2pKa,H3PO4 + 1/2pKa,H2PO4- = 1/2*2.15 + 1/2*7.21 = 4.7

c) Na2HPO4 - both acidic and basic

Na2HPO4 = 2Na+ + HPO42-

Na2HPO4 + NaOH = Na3PO4 + H2O

Na2HPO4 + 2HCl = H3PO4 + 2NaCl

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

d) Na3PO4 - basic

Na3PO4 = 3Na+ + PO43-

PO43- + H2O = HPO42- + OH-

pH = 14 - 1/2pKb,PO43- + 1/2lgCPO43- = 14 - 1/2*1.7 + 1/2lg(0.05) = 12.5

pKb,PO43- = 14 - pKa,HPO42- = 14 - 12.3 = 1.7

e) NH4Cl - acidic

NH4Cl = NH4+ + Cl-

NH4+ + H2O = NH3 + H3O+

pH = 1/2pKa,NH4+ - 1/2lgCNH4+ = 1/2*9.25 - 1/2*lg(0.05) = 5.3

pKa,NH4+ = 14 - pKb,NH3 = 14 - 4.75 = 9.25

f) NaC6H7O7 - both acidic and basic

NaC6H7O7 = Na+ + C6H7O7-

NaC6H7O7 + 2N7aOH = Na3C6H5O7 + 2H2O

NaC6H7O7 + HCl = C6H8O7 + NaCl

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

g) C6H5OH - acidic

C6H5OH = C6H5O- + H+

C6H5OH + NaOH = C6H5ONa + H2O

pH = 1/2pKa,C6H5OH - 1/2lgCC6H5OH = 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 = 1/2pKa,H2CO3 + 1/2pKa,HCO3- = 1/2*6.35 + 1/2*10.32 = 8.3

i) Na2C4H4O4 - basic

Na2C2H2O4 = 2Na+ + CH4O42-

C2H4O42- + 2H2O = C2H6O4 + 2OH-

pH = 14 - 1/2pKb,C2H4O42- + 1/2lgCC2H4O42- = 14 - 1/2*8.36 + 1/2lg(0.05) = 9.2

pKb,C2H4O42- = 14 - pKa,C2H5O2- = 14 - 5.64 = 8.36

j) C4H9O3NH3Cl - acidic

C4H9O3NH3Cl = C4H9O3NH3+ + Cl-

C4H9O3NH3+ = C4H9O3NH2 + H+

pH = 1/2pKa,C4H9O3NH3+ - 1/2lgCC4H9O3NH3+ = 1/2*8.075 - 1/2*lg(0.05) = 4.7

k) CH3NH2 - basic

pH = 14 - 1/2pKb,CH3NH2 + 1/2lgCCH3NH2 = 14 - 1/2*3.34 + 1/2lg(0.05) = 11.7


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Comments

Boye Zephaniah
29.11.19, 17:58

Please the concentration is 0.05M but not 0.5M

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