Question

§Which of the following will not produce a buffered solution?

a.100 mL of 0.1 MNa2CO3and 50 mL of 0.1 MHCl

b.100 mL of 0.1 MNaHCO3and 25 mL of 0.2 MHCl

c.100 mL of 0.1 MNa2CO3and 75 mL of 0.2 MHCl

d.50 mL of 0.2 MNa2CO3and 5 mL of 1.0 MHCl

e.100 mL of 0.1 MNa2CO3and 50 mL of 0.1 MNaOH

I know the answer is e, but I don't understand how the other choices are considered buffers. Aren't buffers supposed to have a Weak Acid/Base with its own salt? (for example, HF with NaF) Isn't HCl a strong acid, how can it make a buffer with MNa2CO3, also how do the Molarities and volumes contribute into determinations of buffers? Please explain thoroughly with a readable handwriting. Thank you.

Answer 1

a.) 100 mL of 0.1 MNa2CO3 and 50 mL of 0.1 MHCl

mmoles of CO32- = 100 x 0.1 = 10

mmoles of HCl = 50 x 0.1 = 5

CO32-   +    H+     ------------>   HCO3-

10               5                                0

5               0                                  5

here HCO3- acid and CO32- conjugate base are remains. so these two can form a buffer.

b.100 mL of 0.1 MNaHCO3 and 25 mL of 0.2 MHCl

mmoles of CO32- = 100 x 0.1 = 10

mmoles of HCl = 25 x 0.2 = 5

CO32-   +    H+     ------------>   HCO3-

10               5                                0

5               0                                  5

here HCO3- acid and CO32- conjugate base are remains. so these two can form a buffer.

c.100 mL of 0.1 MNa2CO3 and 75 mL of 0.2 MHCl

mmoles of CO32- = 100 x 0.1 = 10

mmoles of HCl = 50 x 0.1 = 15

CO32-   +    H+     ------------>   HCO3-

10               15                                0

0              5                                 10

HCO3-   +   H+   ----------> H2CO3

10                5                          0

5                0                           5

here H2CO3 acid and HCO3- conjugate base are remains. so these two can form a buffer.

d.50 mL of 0.2 MNa2CO3 and 5 mL of 1.0 MHCl

mmoles of CO32- = 50 x 0.2 = 10

mmoles of HCl = 5 x 1 = 5

CO32-   +    H+     ------------>   HCO3-

10               5                                0

5               0                                  5

here HCO3- acid and CO32- conjugate base are remains. so these two can form a buffer.

e.100 mL of 0.1 M Na2CO3 and 50 mL of 0.1 M NaOH

here e) is not form a buffer . becuase this is base NaOH