**I. Strong Acids**

The key point is that strong means 100% ionized. For some generic strong acid (with the generic formula HA) dissolved in water, we would write this equation:

HA(s) ---> H^{+}(aq) + A¯(aq)

100% of the HA molecules dissociate in solution. This means we can make the following statement:

The [H^{+}] of a strong acid is equal to the concentration of the acid.

After all, ALL of the acid dissociates. No acid molecules are left.

So, here is a typical problem:

Calculate the pH of a 0.100 M solution of HCl.

In essence, this problem then becomes "Calculate the pH when the [H^{+}] equals 0.100 M." So, to solve it, you write:

pH = −log (0.100) = 1.000

It seems pretty easy and it is. I do have one warning. In the acid base area, there is a LOT of stuff to learn and lots of equations. Usually the idea of strong gets covered near the start of the unit and then doesn't get mentioned much again. What happens is, the kid sees the problem and freaks out because he/she does not recognize it as a strong acid or base. Learn your list of what's strong because everything else you'll see in high school is weak.

Here's another problem:

Calculate the pH of a 1.00 M solution of HBr.

The solution is:

pH = −log (1.00) = 0.00

Yes, that's right, a pH of zero. In fact, you can even have a pH which is negative. It's unusual, but not impossible. Try 2.00 M HCl.

You might want to watch out. A teacher could never mention the possibility of negative pH values in class, but then put one on the test. Ohhhh, those sneaky teachers!

**II. Strong Bases**

Strong bases are pretty much the same as strong acids EXCEPT you'll be calculating a pOH first, then going to the pH. This type of problem is where the relation pH + pOH = 14 is important.

So, the key point is that strong means 100% ionized. For some generic strong base (with the generic formula BOH) dissolved in water, we would write this equation:

BOH(s) ---> B^{+}(aq) + OH¯

100% of the BOH molecules dissociate in solution. This means we can make the following statement:

The [OH¯] of a strong acid is equal to the concentration of the base.

After all, ALL of the base dissociates. No base molecules are left.

So, here is a typical problem:

Calculate the pH of a 0.100 M solution of NaOH.

In essence, this problem then becomes "Calculate the pH when the [OH¯] equals 0.100 M." So, to solve it, you write:

pOH = −log (0.100) = 1.000

pH = 14.000 − 1.000 = 13.000

Here's another:

Calculate the pH of a 0.050 M solution of KOH.

Since the [OH¯] = 0.050 M (remember 100% dissociation), we have this:

pOH = −log (0.050) = 1.30

pH = 14.00 − 1.30 = 12.70

A very brief ending comment:

Generally speaking, weak acids and bases are studied after strong. When teaching about weak acids and bases, the acid dissociation constant (symbol = K_{a}) and the base dissociation constant (symbol = K_{b}) are often introducted. Over the years, I have occasionally had students ask about the values for the K_{a} of a **strong** acid and the K_{b} of a **strong** base.

Here is what I found about K_{a} and K_{b} when applied to strong acids and bases.