I will be doing a lab on Hydrogen Phosphate buffer system tomorrow and I have a few questions about the buffer system. I am not sure if my assumptions are correct.

(1) What would happen to the rate of pH change if 5.0M HCl and NaOH were used instead of 0.50M?

My assumption is that, since the pH of the buffer does not depend on the actual concentration of the buffer, but on the ratio of the two parts, therefore, the rate of pH should not expect to change. Is this correct?

(2) Lets say I prepared a buffer at a different pH, how would the capacity of my buffer system change?

My assumption is that, the buffer capacity would not change upon the change of pH, is this correct?

  • $\begingroup$ +1 Hi Jesse, this is how student questions should be, well done! What I don't understand is if you are adding HCl and NaOH together or if you are using a solution of HCl and then of NaOH to test your buffer. How was the original experiment? $\endgroup$
    – G M
    Mar 25, 2014 at 21:59

1 Answer 1


Let's consider two buffer solutions of the same kind ($pK_a=4.75)$, but of different concentrations:

  • $\ce{CH3COOH}\,0.01\,\text{M}$ and $\ce{CH3COO-} 0.01\,\text{M}$
  • $\ce{CH3COOH}\,0.1\,\text{M}$ and $\ce{CH3COO-} 0.1\,\text{M}$

The starting buffer volume is the same for both and equal to $100\,\text{mL}$.

Now, add a strong base $\ce{NaOH}\,0.05\,\text{M}$, few $\text{mL}$ only: see below $pH$ as a function of $\ce{NaOH}$ added volume:

enter image description here

Have a look how "fast" the $pH$ changes and you will conclude that higher concentrations buffers are more effective. Differences can indeed be appreciated from the graph: they witness the evidence that larger concentrations better "buffer" the addition of the base.

I would suggest you to do some calculations (e.g. for $1\,\text{mL}$ of $\ce{NaOH}\,0.05\,M$ added) and see whether you get the same values of the graph (hopefully I have done them right...). Remember that the addition of a strong base will provoke a decrease in concentration of the acid, $\ce{CH3COOH}$ and a consequent increase of its conjugated base, $\ce{CH3COO-}$.


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