Answer:
na2co3 is a stronger base
na2co3 puts more oh- ion into solution
Explanation:
Step 1: Calculate the pH of 0.1 M Na2CO3

Kb value for Na2CO3 = 2.1 * 10⁻⁴
Set up an ICE table
CO_{3}^{2-} + H_{2} O ------HCO_{3}^{-} + OH^{-} \\[/tex]
I 0.1 - - -
C -x +x +x
E (0.1-x) x x
![Kb = \frac{[HCO3-][OH-]}{[CO3^{2-]} } \\\\Kb = \frac{x^{2} }{(0.1-x)} \\\\2.1*10^{-4} = \frac{x^{2} }{(0.1-x)} \\\\x = [OH-] = 0.00458 M\\\\pOH = -log[OH-] = -log(0.00458) = 2.33\\\\pH = 14 - 2.33 = 11.66](https://tex.z-dn.net/?f=Kb%20%3D%20%5Cfrac%7B%5BHCO3-%5D%5BOH-%5D%7D%7B%5BCO3%5E%7B2-%5D%7D%20%7D%20%5C%5C%5C%5CKb%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%280.1-x%29%7D%20%5C%5C%5C%5C2.1%2A10%5E%7B-4%7D%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%280.1-x%29%7D%20%5C%5C%5C%5Cx%20%3D%20%5BOH-%5D%20%3D%200.00458%20M%5C%5C%5C%5CpOH%20%3D%20-log%5BOH-%5D%20%3D%20-log%280.00458%29%20%3D%202.33%5C%5C%5C%5CpH%20%3D%2014%20-%202.33%20%3D%2011.66)
Step 2: Calculate pH of 0.1M NaHCO3
NaHCO3 is amphiprotic i,e it acts as an acid and base

For amphiprotic systems:
![[H+] = \sqrt{Ka1 * Ka2} \\\\For H2CO3:\\Ka1 = 4.5 * 10^{-7} \\Ka2 = 4.8 * 10^{-11} \\\\[H+] = \sqrt{(4.5*10-7 * 4.8*10-11} = 4.65*10-9\\pH = -log[H+] = 8.33](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%20%5Csqrt%7BKa1%20%2A%20Ka2%7D%20%5C%5C%5C%5CFor%20H2CO3%3A%5C%5CKa1%20%3D%204.5%20%2A%2010%5E%7B-7%7D%20%5C%5CKa2%20%3D%204.8%20%2A%2010%5E%7B-11%7D%20%5C%5C%5C%5C%5BH%2B%5D%20%3D%20%5Csqrt%7B%284.5%2A10-7%20%2A%204.8%2A10-11%7D%20%3D%204.65%2A10-9%5C%5CpH%20%3D%20-log%5BH%2B%5D%20%3D%208.33)
Conclusion:
pH of 0.1 M Na2CO3 = 11.66
pH of 0.1 M NaHCO3 = 8.33
Higher the pH , more basic the solution. Hence, Na2CO3 is a stronger base than NaHCO3