Just say I wanted to give a pH calculation to three significant figures. Eg: the $\mathrm{pH}$ of a $0.500 M$ solution of $\ce{HCl}$ is $-\log(0.5) = 0.3010299957...$

Would 3 s.f. be $0.301$ or $0.30$?

I know normally when zero comes before a number, it isn't considered a significant number (i.e $0.00034$ and $0.34$ both have two significant figures). But since pH is a scale, can I treat $0.30$ as three significant figures?

  • $\begingroup$ Why do you think, it being a pH scale has to use different significant figures? $\endgroup$ – Vedant Oct 23 '16 at 10:34
  • $\begingroup$ you need to use normal convention $\endgroup$ – Vedant Oct 23 '16 at 10:34
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    $\begingroup$ Short answer, no: it doesn't matter that it's a 'scale', you handle its sigfigs exactly as you would any other numerical quantity. Also, beware: never in my experience have I differentiated numerical things by whether they were "a scale" or "not a scale". You may have picked up a false notion there somewhere. $\endgroup$ – hBy2Py Oct 23 '16 at 11:41
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    $\begingroup$ This is a very tricky question, because pH is a log value. So, only the mantissa means anything in terms sig figs. You really need error bars, but I would go with 2. Importantly, note that 4.30 pH has the same sig figs as 0.30 pH. $\endgroup$ – Zhe Oct 23 '16 at 12:59
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    $\begingroup$ I hope my newly edited answer helps. I also come to the conclusion that 2 decimal places are fine assuming that the error is $\pm$ 0.005 M : ) $\endgroup$ – logical x 2 Oct 23 '16 at 17:38

If 0.500 means "between 0.495 and 0.505" then

-log[HCl] is between 0.305 and 0.296

so statistically 3 digits beyond the decimal point should be used. The uncertainty in the third digit is almost identical in both values ([HCl] and -log[HCl]).

Systematic error from approximating pH as -log[HCl] will be much greater. 0.3 is wrong by more than 0.1 pH units because the activity coefficient of 0.5M (H+, Cl-) is about 0.75.


Short Answer:

The number of significant figures in the mantissa of a value expressed in scientific notation equals the number of significant figures to the right of the decimal in the logged value. So in your example, assuming the likely error is the last digit, I'd use two.

Now back to your question, you have $[\ce{HCl}]=0.500$ assuming, the error is in the last digit, I'd recommend using 2 significant figures, $\pm$ error in the last.

Additionally, just because pH is a scale doesn't mean you can treat $0.30$ as a number with three significant figures, it still has two.

The general principles are given below. Concepts like this are best illustrated with an example.

Consider, a random number $0.0000273$ Let us write this in scientific notation for the sake of clarity, $0.0000273 = 2.73×10^{–5}$ We have three significant figures.

Now take the log on both sides

$\log(0.0000273)= \log(2.73×10^{–5}) = \log(2.73) + \log(10^{–5})$

Now let us examine the two terms separately,

First, $\log(2.73) = 0.436$: the answer has three significant figures, reflecting the possible error in the last digit of $2.73$

$\log(10^{–5}) = -5$ This is an exact number and has infinite significant figures, i.e there is no error.

Going back, $\log(0.0000273)= \log(2.73) + \log(10^{–5}) = 0.436 -5.0000...$

In this situation the rules for significant digits for sums, dictate that we can only add to the same decimal place as the value with the least significance. In this case, that is the third decimal place from 0.436, so

$\log(0.0000273)= \log(2.73) + \log(10^{–5}) = 0.436 -5.0000... = -4.564$

$\log(0.0000\overline{273}) = -4.\overline{564}$ (the overline indicates the 3 significant figures)

End of example. Now back to your question, you have $[\ce{HCl}]=0.500$ assuming, the error is in the last digit, I'd recommend using 3 digits after the decimal, $\pm$ error in the last. So, $0.301 \pm 0.00\epsilon$

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    $\begingroup$ Answer is saying that OP has "pH=0.500", but instead [HCl]=0.500. $\endgroup$ – DavePhD Oct 23 '16 at 19:03
  • $\begingroup$ @DavePhD Thanks for your comment, sorry my mistake. Careless, careless me $\endgroup$ – getafix Oct 23 '16 at 19:11
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    $\begingroup$ thank you so much for all the time and effort you must have put into this great answer! $\endgroup$ – Christy Oct 24 '16 at 10:09

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