## Chemistry Tutorial

*Some pH, pOH, [H+], [OH-], and Kw Relationships*

#### Intro

When dealing with acid/base chemistry some of the quantities can seem difficult to comprehend and even redundant. Try your hand at some of these relations!

#### Sample Problem

Express these quantities using the others:

Keeping in mind that [H+] * [OH-] = Kw = 10^-14

#### Solution

Some definitions first:

pH = -log[H+] and pOH = -log[OH-]

Math concepts:

if y = log(x), then x = antilog(y), where antilog(y) is merely 10^y

log(x) + log(y) = log(xy)

Using these definitions and some algebraic manipulation (multiplying both sides by -1), we get: -pH = log[H+] and -pOH = log[OH-], this might seem trivial but by undoing the log we can directly solve for [H+] and [OH-] by taking the antilog, giving us:

[H+] = antilog(-pH) = 10^-pH and [OH-] = antilog(-pOH) = 10^-pOH

Next we can further exploit logarithms even more:

We know [H+] * [OH-] = 10^-14 and by taking -log of both sides we get:

-log([H+] * [OH-]) = -log(10^-14) and by using the mathematical concept above we get:

-log[H+] + -log[OH-] = -log(10^-14) and by using definitions above and solving the right hand side log we get: pH + pOH = 14

So we now have knowledge that:

pH = -log[H+] and pOH = -log[OH-]

[H+] = antilog(-pH) = 10^-pH and [OH-] = antilog(-pOH) = 10^-pOH

[H+] * [OH-] = 10^-14 –> [H+] = (10^-14)/[OH-] and [OH-] = (10^-14)/[H+]

pH + pOH = 14 –> pH = 14 – pOH and pOH = 14 – pH

I hope this was clear enough, and that some of you get some milage out of seeing all these quantities in one place.

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