Chem Problem, Buffers?
.40g of solid NaOH is added to 1 liter of a buffer sol. that is .1M in CH3COOH and .1M in NaCH3COO, how will the pH of the soln. Change.
I came out with the answer that it doesnt change? Anyone disagree or agree?
Answer:
pH = pKa + (log base)/(log acid)
pH = 4.76 + (log .1)/(log .1)
pH = 4.76 originally
Then we add the NaOH and it changes.
We added exactly 1 mol of NaOH based on the molar mass, and in 1 liter of solution that adds 1.0 M base.
So the new equation is:
pH = 4.76 + (log1.1)/(log.1)
pH = 4.76 - .0414
pH = 4.72
So it barely changes. You were correct
Hope that helps
The pH will increase. NaOH is a strong base. Acetic acid and its conjugate base are very weak. Since the NaOH is being added, than the pH will be higher because the solution is becoming more basic.
I would agree. The pKa of acetic acid is 4.76 therfore
pH = pKa + log[base]/[Acid]
pH = 4.76 + log(1)
pH = 4.;76 + 0
pH = 4.76
Also the reason why they call them buffers is that the pH will not change with the addition of an amount of acid or base. It basically all depends on the buffering capacity The buffering capacity is the ability of a buffer to resist changes in pH.. The maximum buffering capacity is obtained when the pH =pKa. The closer the buffered pH is to the pKa of the acid the greater the buffering capacity. Therfore you would have to add a good amount of acid or base to even touch the pH.
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