[plofa]

Equation is Na2CO3 + H20 = NaHCO3 + NaOH

The Questions
a)A solution of 0.2 M Na2CO3 (pKa = 10.33) is prepared. What is its pH?
--for this i used pH = 0.5pKW + 0.5pkA + 0.5log(base) so i got 11.82 as the pH

b) 2.52 g of NaHCO3 (MW = 84.0; pKa = 10.33) was added to 100 ml of the
solution prepared in (b) above. What is the pH of the resultant buffer solution?

c)To the buffer solution prepared in (c) above, 0.5 ml of 2 M NaOH was added.
What is the change in pH of the buffer solution?



I don't get how u do c and d i know you use the equation pH= pKa + log(base/acid)

Thanks any help would be greatly appreciated!

[Foobarz]

OK
If I am not mistaken

the 100 mL soln has 0.03 mol HCO₃^-
and 0.02 mol CO₃^2-

and when you add 0.5 mL of 2 M NaOH you're adding 0.001 mol OH^-.

Ice table time:

       HCO₃^- + OH^-   CO₃^2- + H2O
I      0.03      0.001    0.02        
C    -0.001    -0.001    +0.001
E     0.029      0         0.021

And 0.029 mol HCO3 /(100+0.5 mL) = 0.289 M HCO3
And 0.021 mol CO3 / (100+0.5 mL) = 0.209 M CO3

and with those new molarities you can do your henderson hasselbalch stuff
I got a new pH of 10.19

[Borek]

In such problems you just assume neutralization went to completion. As long as acids/bases involved are not too strong/weak/diluted it works OK.

[Foobarz]

Well, I think I am correct as the answer in part b, the pH is 10.15.

And for part C, doing Henderson Hasselbalch I get a pH of 10.19.

Which can be expected for putting just a teensy, 0.5 mL of NaOH into 100 mL of a buffer soln that resists large pH changes.