Question

Compare the ph of 0.1 m na2co3 and 0.1 m nahco3. select the statements that explain the difference. na2co3 is a stronger base. nahco3 is a stronger base. na2co3 puts more oh- ion into solution.

Answer 1

Na₂CO₃ is a stronger base while
Na₂CO₃ puts more OH- ion into solution
The strongest base is [H₊]= 10 - 12
The strongest acid is PH = 0.5
If an acid is added to neutral water the H₊
will increase but when an acid added to neutral water 
the OH⁻ will decrease.

Answer 2

Answer:

na2co3 is a stronger base

na2co3 puts more oh- ion into solution

Explanation:

Step 1: Calculate the pH of 0.1 M Na2CO3

$Na_{2} CO_{3} ----- 2Na^{+} + CO_{3} ^{2-} \\\\CO_{3}^{2-} + H_{2} O ------HCO_{3}^{-} + OH^{-}$

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

$Kb = \frac{[HCO3-][OH-]}{[CO3^{2-]} } \\\\Kb = \frac{x^{2} }{(0.1-x)} \\\\2.1*10^{-4} = \frac{x^{2} }{(0.1-x)} \\\\x = [OH-] = 0.00458 M\\\\pOH = -log[OH-] = -log(0.00458) = 2.33\\\\pH = 14 - 2.33 = 11.66$

Step 2: Calculate pH of 0.1M NaHCO3

NaHCO3 is amphiprotic i,e it acts as an acid and base

$NaHCO3 ------- Na+ + HCO3-\\\\HCO3- ------ H+ + CO3^{2-} (acts as acid, H+ released)\\\\HCO3= + H2O ------- H2CO3- + OH- (acts as base, OH- released)$

For amphiprotic systems:

$[H+] = \sqrt{Ka1 * Ka2} \\\\For H2CO3:\\Ka1 = 4.5 * 10^{-7} \\Ka2 = 4.8 * 10^{-11} \\\\[H+] = \sqrt{(4.5*10-7 * 4.8*10-11} = 4.65*10-9\\pH = -log[H+] = 8.33$

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