Question

Consider the following acids and bases: at HCO,H HOBI K = 1.8 x 104 Kg = 2.0 x 10-9 (C2H3)2NH Kb = 1.3 x 10-3 HONH2 Kb = 1.1 x 10-8 Choose substances from the following list that would be the best choice to prepare a pH = 9.0 buffer solution. (Select all that apply.) OCH3)2NH2C1 HOBI (CH3)2NH NaOBI KHCO2 HONH OHCOH OHONH2NO3 Submit

Answer 1

Ans-

HOBr, NaOBr,

Explanation-

A Buffer is a solution where both a weak acid (HA) and its conjugate base (A-) are presnet or both a weak base (HB) and its conjugate acid (H2B+) are presnet

Now a buffer is effective when the pH of the buffer comes in a range on -1 of pKa to +1 of pKa.

So to have a buffer of pH 9, we have to see for which set the pH value comes betwen -1 of pKa to +1 of pKa for the given species-

So on this basis, lets calculatae the pKa and pKb for the species directly-

a-

HCO2H

pKa = -log Ka = -log ( 1.8 * 10-4) = 3.74

So buffer with pH = 9 can't be prepared from this acid or the salt of this species (Ex- KHCO2, HCO2H)

b- HOBr (Ka = 2.0 * 10-9 )

So

pKa = -log Ka = -log ( 2.0 * 10-9) = 8.7

So possible pH for buffer = -1 of pKa to +1 of pKa i.e from 7.7 to 9.7

So buffer with pH = 9 can be prepared from this acid or the salt of this species (Ex- HOBr, NaOBr)

c- (C2H5)2NH (Kb= 1.3 * 10-3 )

So

pKb = -log Kb = -log (1.3 * 10-3) = 2.9

So

pKa = 14 - pKb = 14 - 2.9 = 11.1‬

So possible pH for buffer = -1 of pKa to +1 of pKa i.e from 10.1 to 12.1

So buffer with pH = 9 can't be prepared from this acid or the salt of this species (Ex- (C2H5)2NH, (C2H5)2NH2Cl)

d-

HONH2 (Kb= 1.1 * 10-8)

So

pKb = -log Kb = -log (1.1 * 10-8) = 8

So

pKa = 14 - pKb = 14 - 8 = 6‬

So possible pH for buffer = -1 of pKa to +1 of pKa i.e from 5 to 7

So buffer with pH = 9 can't be prepared from this acid or the salt of this species (Ex- HONH2, HONH3(NO3))

Examples of some calculation-

i- (C2H5)2NH2Cl ----------> (C2H5)2NH2+ + Cl-  

Here the species (C2H5)2NH2+ act as a weak acid which can form the conjugate base as-

(C2H5)2NH2+ -----------> (C2H5)2NH + H+   

Given Kb for (C2H5)2NH = 1.3 * 10-3  

So

pKb = -log Kb = -log (1.3 * 10-3) = 2.9

So

pKa = 14 - pKb = 14 - 2.9 = 11.1‬

So possible pH for buffer = -1 of pKa to +1 of pKa i.e from 10.1 to 12.1

So our required buffer is pH= 9 can't be formred from this.

ii-

HOBr (Ka = 2.0 * 10-9 )

So

pKa = -log Ka = -log ( 2.0 * 10-9) = 8.7

So possible pH for buffer = -1 of pKa to +1 of pKa i.e from 7.7 to 9.7

So our required buffer is pH= 9 can be formred from this.

iii-

Kb for (C2H5)2NH = 1.3 * 10-3  

So

pKb = -log Kb = -log (1.3 * 10-3) = 2.9

So

pKa = 14 - pKb = 14 - 2.9 = 11.1‬

So possible pH for buffer = -1 of pKa to +1 of pKa i.e from 10.1 to 12.1

So our required buffer is pH= 9 can't be formred from this.

iv-

NaOBr

NaOBr ----------> Na+ + OBr-  

Here the species OBr- act as a weak base which can form the conjugate acid as-

OBr​​​​​​​- + H2O -----------> HOBr + OH-    

Now we have pKa for HOBr is 2.0 * 10-9

So

pKa = -log Ka = -log ( 2.0 * 10-9) = 8.7

So possible pH for buffer = -1 of pKa to +1 of pKa i.e from 7.7 to 9.7

So our required buffer is pH= 9 can be formred from this.