For a tube with 1 mL of 0.75 % (w/v) casein in 0.10 M sodium acetate, with 0.05 mL acetic acid, and 6.95 mL H2O, how do you determine the pH?

My original thought was to use the Henderson-Hasselbalch equation with the concentration of base being 1 and the concentration of acid being 0.05, but this does not feel right.

I think my fundamental understanding of the direct correlation between concentration and volume is not correct.

  • $\begingroup$ Is the concentration of the original 0.05 ml acetic acid given in your exercise? $\endgroup$ – Loong Sep 17 '16 at 11:26
  • $\begingroup$ Sorry, the concentration of acetic acid is also 0.10M $\endgroup$ – Sushifish Sep 17 '16 at 11:49

With regard to your understanding of the correlation between concentration and volume, you should know that concentration $c$ is defined as $$c=\frac nV$$ where $n$ is amount of substance and $V$ is the volume of the solution.

When you dilute a sample, you change the volume $V$, but you do not change the amount of substance $n$ in your sample. Therefore, if you know the initial concentration $c_1$ in the initial volume $V_1$ $$c_1=\frac n{V_1}$$ you can calculate the new concentration $c_2$ after dilution to the new volume $V_2$: $$c_2=\frac n{V_2}$$ You can rearrange both equations to calculate the amount of substance $n$: $$\begin{align} c_1\cdot V_1&=n\\[6pt] c_2\cdot V_2&=n \end{align}$$ Since the amount of substance $n$ is constant, you get another useful equation: $$c_1\cdot V_1=c_2\cdot V_2$$

In your exercise, after addition of all solutions, the new volume may be estimated as $$V_2\approx1\ \mathrm{ml}+0.05\ \mathrm{ml}+6.95\ \mathrm{ml}=8\ \mathrm{ml}$$

The initial concentration of the base (sodium acetate) is $c_{\text{base},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{base},1}=1\ \mathrm{ml}$.

Therefore, the new concentration of the base is

$$\begin{align} c_{\text{base},1}\cdot V_{\text{base},1}&=c_{\text{base},2}\cdot V_2\\[6pt] c_{\text{base},2}&=\frac{c_{\text{base},1}\cdot V_{\text{base},1}}{V_2}\\[6pt] &=\frac{0.10\ \mathrm{mol\ l^{-1}}\times1\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt] &=0.0125\ \mathrm{mol\ l^{-1}} \end{align}$$

The initial concentration of the acid (acetic acid) is $c_{\text{acid},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{acid},1}=0.05\ \mathrm{ml}$.

Therefore, the new concentration of the acid is

$$\begin{align} c_{\text{acid},1}\cdot V_{\text{acid},1}&=c_{\text{acid},2}\cdot V_2\\[6pt] c_{\text{acid},2}&=\frac{c_{\text{acid},1}\cdot V_{\text{acid},1}}{V_2}\\[6pt] &=\frac{0.10\ \mathrm{mol\ l^{-1}}\times0.05\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt] &=0.000625\ \mathrm{mol\ l^{-1}} \end{align}$$

Finally, you can use the Henderson–Hasselbalch equation to calculate the $\mathrm{pH}$ of the acetate buffer from the new concentrations $c_{\text{acid},2}$ and $c_{\text{base},2}$.

  • $\begingroup$ Thank you very much for this, this helps a lot with my understanding of the actual change that is taking place concentration wise. My only question is that performing this calculation with these new concentrations, or with the old concentrations result in the same pH yield. Am I to understand that this is because the ratio of base to acid is still the same? Does this mean that the addition of water has no effect on pH at least according to the equation used? $\endgroup$ – Sushifish Sep 17 '16 at 12:41
  • 1
    $\begingroup$ @Sushifish Yes, if the assumed volumes for the acid and the base are identical, this volume cancels out in the Henderson–Hasselbalch equation. $\endgroup$ – Loong Sep 17 '16 at 12:57

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