Suppose we want to calculate pH of an NHX4HCOX3 solution of known concentration. Now, NHX4X+ will be hydrolised to give HX+ ions. Some HCOX3X− ions will be converted to (i) HX2COX3 and OHX− and some will give (ii) HX+ and COX3X2−
HCOX3X−+HX2O−⇀↽−HX2COX3+OHX−1/Ka1HCOX3X−+HX2O−⇀↽−COX3X2−+HX3OX+Ka2NHX4X++HX2O−⇀↽−NHX3(aq)+HX3OX+1/Kb
I am not sure how to take account of both HX3OX+ and OHX− in the calculation of pH.
Will the third equilibium costant be 1/Kb?
Answer
The main reactions which take place are
1)Ionization of HAX−: HAX−⟶HX++AX2−
2)Hydrolysis of HAX−: HAX−+HX+⟶HX2A
3)Hydrolysis of cation: BX++HX2O⟶BOH+HX+
for reaction 1, ka2=[H+][A2−][HA−]
for reaction 2, ka1=[HA−][H+][H2A]
for reaction 3, kh=kwkb=[BOH][H+][B+]
At any moment, [HX+]=[BOH]+[AX2−]−[HX2A]
Now we will have to make 2 assumptions. First lets assume that ka1<<[HCO−3] so that ka1+[HCO−3]=[HCO−3]
[H+]=√ka1⋅([NH+4].kwkb+ka2[HCO−3])[HCO−3]
If you assume that neither hydrolysis nor the dissociation goes too far, you can assume the concentration of bicarbonate ion to be equal to the initial concentration (say C). Since Ammonium bicarbonate dissociates to give two products in 1:1 ratio, concentration of ammonium produced is equal to the concentration of bicarbonate ion produced.
I saw this formula for calculation pH of such salts in my textbook but no derivation was given: [H+]=√ka1⋅(kwkb+ka2)
The above formula can be obtained if you consider the last assumption. (Clarification needed)
No comments:
Post a Comment