My question is:
I want to do a buffer with benzoic acid and NaOH, and I will call this buffer when its finish for “Solution A”.
I want to have 200 ml of the buffer, and I have 1,32 g benzoic acid and 24 ml of 1,00 M NaOH.
C6H5COOH + OH- C6H5COOH + H2O
1. first i want to calculate the buffer pH, and i have done like this:
Vtot= 200*10-3ml
mbenzoic=1,32 g
Mwbenzoic=122g/mol
nbenzoic= 0,01 mol
VNaOH= 24*10-3 ml
CNaOH= 1 M
nNaOH= 0,024 mol
Then I calculated the concentration in 200 ml:
CNaOH= n/V => 0,024/200*10-3 = 0,12 M
Cbenzoic= n/V => 0,01/200*10-3= 0,05 M
Now I will use Henderson–Hasselbalch equation..
And the pKa for benzoic acid is 4,19.. So with the Henderson–Hasselbalch equation i will get the pH at the buffer: 6,59.
Have i thinking right now?
2. My other question is that i want to calculate pH if I take 20 ml “Solution A” (the solution above) and establish 1ml, 1M NaOH. How do I calculate this pH?
Is it the concentration for benzoic acid that I use for the calculation?