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.
\mathrm{
and}
. I think for logarithms you can also use\log
. $\endgroup$ – Jan Sep 24 '15 at 12:06