Question

A buffer is created by combining 160 mL of 0.20 M HCHO2 with 75 mL of 0.25 M NaOH. Determine
the pH of the buffer.
Someone help me to figure out this question..

Answer 1

Nyte it wouldn't help you much posting this here. I suggest you to post this in Chemistry, you'd get better response there. Thanks :)

Answer 2

ok.. :)

Answer 3

HCHO2 is your weak acid, NaOH is your strong base.

Answer 4

HCHO2 + NaOH ---> Na+HCO2- + H20

Therefore, the pKa we're looking for is that of sodium bicarbonate (the one to the left of H20).

The pKa is 10.329 http://en.wikipedia.org/wiki/Sodium_bicarbonate

Answer 5

Now for your concentrations. You're using Henderson-Hasselbalch equation:

pH = pKa + log ([base]/[acid])

Total volume of buffer solution: 235 mL

Moles of HCHO2: 160 mL * (0.2 mols/L) = 32 mols HCHO2

Concentration of HCHO2: (32 mols HCHO2)/(235 mL) = 0.136 M


Mols NaOH: 75 mL * (0.25 mols/L) = 18.75 mols

Concentration of NaOH: (18.75 mols NaOH)/(235 mL) = 0.0798 M

Okay so in summary, we have:
pKa = 10.329
[acid] = 0.136 M
[base] = 0.0798 M

pH = 10.329 + log (0.0798/0.136)

Answer 6

eh, who am I kidding. I'm not sure if it's right but I tried :-)