Significant figures in pH calculations: does pH 0.30 count as 2 or 3 significant figures?
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, $0.301 \pm 0.00\epsilon$