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I have just started learning how to calculate pH and I have noticed that adding [$\ce{H+}$] makes a negligible difference in few cases whereas in other cases it makes a major change in the final value. Why is that?

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    $\begingroup$ Can you provide an example of what you mean? $\endgroup$ – Zhe Aug 8 '17 at 13:59
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pH is calculated by a logarithmic formula given by:

$$\text{pH}=-\log{\ce{[H+]}}$$

Lets say, the initial hydrogen ion concentration is $\ce{[H+]_0}$, and change in concentration is $d\ce{[H+]}$. The change in pH would be:

$$d\text{pH}=-\log{(\ce{[H+]_0}+d\ce{[H+]})}$$

This can otherwise be written as:

$$d\text{pH}=-\log{\ce{[H+]_0}}×\frac{d\ce{[H+]}}{\ce{[H+]_0}}$$

You can see that the change in pH depends on the initial hydrogen ion concentration. It's in fact inversely proportional to the initial concentration. Adding a drop of concentrated acid to a base would give a much bigger change in pH compared to adding a drop to pure water, which in turn is bigger compared to adding a drop to an acid.

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  • $\begingroup$ So do I need to include the hydrogen ion concentration while calculating pH ? $\endgroup$ – Justuraveragemathsstudent May 10 '17 at 12:10
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    $\begingroup$ @Justuraveragemathsstudent Of course! pH itself is a measure of hydrogen ion concentration. $\endgroup$ – Pritt Balagopal May 10 '17 at 12:13
  • $\begingroup$ sorry..I meant hydrogen ion concentration of water $\endgroup$ – Justuraveragemathsstudent May 10 '17 at 12:14
  • $\begingroup$ @Justuraveragemathsstudent Well, it depends on the strength of the acid, and how much of it is there. If the acid produces much more ions than water does, ignore it, or else take it into consideration. $\endgroup$ – Pritt Balagopal May 10 '17 at 12:18
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    $\begingroup$ I don't follow the algebra here the change in pH is $\delta pH=-\log(H_0)+\log(H_0+\delta H)$, rearranging gives $\delta pH=-\log(H_0)+\log(H_0[1+\delta H/H_0])$ which becomes after expanding and then expanding the log with $\delta H/H_0 \lt 1$ and taking just the first term $\delta pH=\delta H/H_0$. $\endgroup$ – porphyrin May 10 '17 at 14:03

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