Quick run down, I was given pH of buffer 4.5 and then pKa of acid 4.79 I then had to work out volume of acid and volume of base using $\text{pH=pKa}+\log_{10}{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$. Using that I calculated the volumes of acid and base required to make 15 ml of the buffer $\ce{CH3COOH,CH3COONa}$:

$V_{\ce{HA}}=9.7\ \text{ml}$ and $V_{\ce{A-}}=5.3\ \text{ml}$ (both acid and base used to make buffer are of 0.1 M conc, probably come in handy later).

Then I am now asked to calculate the amounts of acid and base in 15ml of my buffer they give me Acid: $n_{\ce{HA}} = [\ce{HA}]\cdot V_{\ce{HA}}=0.1\cdot V_{\ce{HA}}$

Base: $n_{\ce{A-}} = [\ce{HA}]\cdot V_{\ce{A-}}=0.1\cdot V_{\ce{A-}}$

Which I don't really get? I then have to find out the amount of acid was added, some context is that this is n experiment I conducted using a pH probe I added three drops and recorded the pH for a total of 60 drops the final pH reading is 3.52 of an initial reading of 4.38, would I use the simple $\text{pH} = -\log_10[\ce{H+}]$? They have given me $[\ce{H^+}]\cdot V=0.1\text{ M}\cdot\dfrac{1\text{ ml}}{1000\text{ L}}=...\text{mol}$ which I also don't really know what to do with it. After that I calculate the new concentrations of acid and conjugate base after all added H+ reacts initial and final. you don't have to use my example just explain EVERY SINGLE STEP and guide me through the process :) much appreciated! I know I am asking a lot in one go but this all correlates with one another, if anyone can help thanks a lot!

  • $\begingroup$ Could you include the whole question (or a summary)? I'm having trouble understanding what's going on and where $\dfrac{[A^-]}{[HA]} = 0.5495$ come from. $\endgroup$
    – Jerry
    May 11, 2013 at 14:50
  • $\begingroup$ I have given a brief summary hope that makes more sense! Though I think its cause I don't actually know what to do which makes my question poorly worded. $\endgroup$ May 12, 2013 at 2:20

1 Answer 1


I'm going to address only the first part since that was in the original question.

You were given a buffer of pH 4.5 and the pKa of the acid used is 4.79. You worked out the volumes of the acid to be 9.7 mL and the volume of base to be 5.3 mL. You are now asked to get the amount of acid and the amount of base (i.e. the number of moles of each in the buffer).

$n_a$ means the number of moles of the acid.

This is equal to the concentration of the acid multiplied by the volume of the acid, namely:

$n_a = [\ce{HA}].V_a$

Since we know that the concentration of the acid, $[\ce{HA}]$ is 0.1 M and the volume of acid used is 9.7 mL, the number of moles of acid becomes:

$n_a = 0.1\ \text{mol/L}\times 9.7\ \text{mL} \times\dfrac{1\ \text{L}}{1000\ \text{mL}}$

The $\dfrac{1\ \text{L}}{1000\ \text{mL}}$ is only to convert mL to L. Do the same for the number of moles of the conjugate base.

For the second part, you need to realise that the pH you get that way will be the combined pH of the buffer and the acid. You will probably have to use the assumption that 1 drop = 1 mL, unless the question or your tutor already mentioned how to consider the volume of the added acid.


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