The question in my class notes read as:

Calculate the p-value for each ion in a solution that is $2.00 \times 10^{-7}\ \mathrm{M}$ in $\ce{NaCl}$ and $5.4 \times 10^{-4}\ \mathrm{M}$ in $\ce{HCl}$.

I was confused how to approach it, but my teacher did the solution on the board with the work done out and obtained the values of $\mathrm p\ce{H}$, $\mathrm p\ce{Na}$ and then used the totals of those values to get the concentration of [$\ce{Cl-}$]. From that, she then took the negative log of the concentration of [$\ce{Cl-}$].

For reference, she obtained the following values: $$\begin{align} \mathrm p\ce{H} &= 3.27, \\ \mathrm p\ce{Na} &= 2.699, \\ [\ce{Cl-}] &= 2.54 \times 10^{-3}, \\ \mathrm p\ce{Cl} &= 2.595 \end{align}$$

I understand how one has to take the negative log of a given concentration to obtain the p-value of it in a "rote" type of sense.

But what I misunderstood was the log rules and the calculation of the first $\mathrm p\ce{H}$ value.

Her work for the calculation of $\mathrm p\ce{H}$ was as followed:

(1st step)

$\mathrm p\ce{H} = -\log[\ce{H3O+}] = -\log[5.4 \times 10^{-4}]$

(use the log rule, $\log(a \cdot b) = \log(a) + \log(b)$)

(2nd step)

$\mathrm p\ce{H} = -\log 5.4 - \log 10^{4}$

(3rd step)

$\mathrm p\ce{H} = 3.27$

But why in the 2nd step above is the $\log 10^{4}$ not have a negative value, such as $\log 10^{-4}$?

In her solution of p$\ce{Na}$ it was the case that the second additive had a negative exponent, which can be seen below in the second step.

(1st step)

$\mathrm p\ce{Na} = -\log[2.00 \times 10^{-3}]$

(2nd step)

$\mathrm p\ce{Na} = -\log2.00 + \log10^{-3}$

(3rd step)

p$\ce{Na}$ $= -0.301 -(-3.00) = 2.699$

Also if anyone could explain to be why we sum the values of $\mathrm p\ce{H}$ and $\mathrm p\ce{Na}$ to obtain $[\ce{Cl-}]$? Does this have to do with the "stoichiometry" of something? My chemistry is not great, nor is my math. Thank you.

  • 1
    $\begingroup$ You should format both units and functions (such as M and log) in upright type in maths mode. This always works by enclosing them with \mathrm{ and }. I think for logarithms you can also use \log . $\endgroup$
    – Jan
    Commented Sep 24, 2015 at 12:06
  • $\begingroup$ Also to answer the first question: Yes, that is a typing error. It should be $-\log \left (2.0 \cdot 10^{-3} \right ) = - \left (\log 2.0 + \log 10^{-3} \right ) = - \log 2.0 - \log 10^{-3} = - \log 2.0 + \log 10^3$ $\endgroup$
    – Jan
    Commented Sep 24, 2015 at 12:10
  • $\begingroup$ @Jan So would: $pH = - \log{5.4 * 10^{4}}$, and then $pH =−\log{5.4} − \log{10^{4}} $ be correct or no? $\endgroup$
    – Ro Siv
    Commented Sep 24, 2015 at 14:08
  • $\begingroup$ Technically the calculation is correct but not realistic. However: $p\ce{H} = - \log 5.4 \cdot 10^{-4} = - \log 5.4 - \log 10^{-4} = - \log 5.4 + \log 10^4$ $\endgroup$
    – Jan
    Commented Sep 24, 2015 at 14:29

1 Answer 1


It seems like the calculation done or copied from the board by you has gone a little wrong.

In your first step when taking $-\log$ of the given concentration you have ignored the signs on the lower of the argument of the log function. $$-\log 10^{-3} = \log 10^3 = 3 \log 10 = 3$$ since the base of log here is 10. And same correction for next steps.

And to your other question ... We add up the concentrations (adding up p values is irrevalent) because here NaCl is a strong electrolyte and HCl is a strong acid. They both dissociate completely and individually contribute to the concentration of chlorine anion.

  • $\begingroup$ Corrected , was a mistake . I'm sorry. $\endgroup$ Commented Sep 24, 2015 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.