p-value for ions in a solution?

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.




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.


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