Question

1)An aqueous solution with a pH of 2.78 is diluted from 1.79 L to 3.34 L. What is the pH of the diluted solution?

2)An oxalic acid (H₂C₂O₄) solution has the following species present in varying concentrations. Which of these species can act as a base?

1. H₂C₂O₄

2. HC₂O₄^–

3. C₂O₄^2–

4. H₃O⁺

Answer 1

1) pH = 2.78 , [H+] = 10^ -pH = 10^ -2.78 = 0.00166 M

now when we deal with dilution we use formula M1V1 = M2V2

where M1 = 0.00166 M , V1 = 1.79 L , V2 = 3.34 L , we find M2

0.00166 x 1.79 = M2 x 3.34

M2 = 8.894 x 10^ -4 M

pH = -log [H+]= -log ( 8.894x10^-4) = 3.05

2) H2C2O4 (aq) <---> HC2O4 -(aq) + H+ (aq)

hence HC2O4- can act as base by absorbing H+ to fornm oxalic acid

similarly HC2O4- (aq) -<----> C2O42-(aq) + H+ (aq)

Hence C2O4 2-(aq) can also act bas base

Hence HC2O4-(aq) and C2O4 ^2-(aq) can act as base