Calculate the pH of a solution that is 2.00M HF, 1.00M NaOH, and 0.393M NaG (Ka=7.2 x10^-4).
The answer I got was 4.76 ..
3 answers
What in the world is NaG? Could that be NaF? If so, did you use Henderson-Hasselbalch equation and adjust it for the 1.00 M NaOH? And how much NaOH was added.
Sorry typo..it's NaF and it doesn't say how much was added that's the question given
I would do this.
2.00M HF = 1.00M NaOH gives 1.00M NaF and leaves 1.00 M HF.
1.00N NaF + 0.393M = 1.393M
pH = pKa + log(base)/(acid)
I get something like 3.28.
2.00M HF = 1.00M NaOH gives 1.00M NaF and leaves 1.00 M HF.
1.00N NaF + 0.393M = 1.393M
pH = pKa + log(base)/(acid)
I get something like 3.28.
2 answers