# 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.

• 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 . – Jan Sep 24 '15 at 12:06
• Also to answer the first question: Yes, that is a typing error. It should be $-\log \left (2.0 \cdot 10^{-3} \right ) = - \left (\log 2.0 + \log 10^{-3} \right ) = - \log 2.0 - \log 10^{-3} = - \log 2.0 + \log 10^3$ – Jan Sep 24 '15 at 12:10
• @Jan So would: $pH = - \log{5.4 * 10^{4}}$, and then $pH =−\log{5.4} − \log{10^{4}}$ be correct or no? – Ro Siv Sep 24 '15 at 14:08
• Technically the calculation is correct but not realistic. However: $p\ce{H} = - \log 5.4 \cdot 10^{-4} = - \log 5.4 - \log 10^{-4} = - \log 5.4 + \log 10^4$ – Jan Sep 24 '15 at 14:29

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.