**Step 1:**

Determine if the acid is monoprotic, diprotic, or triprotic. This will help you determine the concentration of the H^{+}. If the acid is monoprotic, or has 1 hydrogen in the front of the formula, the concentration of the H^{+} is the same as the concentration of the acid.

Example: 2.5 M HCl has a concentration of H^{+} = 2.5 M

If the acid is diprotic, or has 2 hydrogens in the front of the formula, the concentration of the H^{+} is two times the concentration of the acid.

Example: 2.5 M H_{2}SO_{4} has a concentration of H^{+} = 2 x (2.5 M) = 5.0 M

If the acid is triprotic, or has 3 hydrogens in the front of the formula, the concentration of the H^{+} is three times the concentration of the acid.

Example: 2.5 M H_{3}PO_{4} has a concentration of H^{+} = 3 x (2.5 M) = 7.5 M

Once you have determined the concentration of H^{+} (also denoted as [H^{+}]), you can move to step 2.

**Step 2:**

Using the formula for pH (pH = -log [H^{+}]), plug in your value for [H^{+}].

**Step 3:**

Using your scientific calculator, calculate the value for pH.

**Determine the pH of a 0.010 M HNO**_{3} solution.

_{3}solution.

**Step 1:**

Because there is only 1 H at the beginning of the formula of this acid, it is monoprotic. Therefore, [H^{+}] = the concentration of the acid.

[H^{+}] = 0.010 M or 1.0 x 10^{-2} M

**Step 2:**

pH = -log [H^{+}] = -log [1.0 x 10^{-2}]

**Step 3:**

pH= -(-2.0)

pH = 2.0

Note: Scientific calculators all work differently. You need to learn how your calculator works, and always use the same calculator for all assignments, quizzes, and tests. Try this problem until you determine the order of entry that gives you the correct answer. Notice that you are taking the negative log. When you type log 1.0 EE -2 into your calculator and hit equals, you will see -2.0 displayed. Multiply this answer by -1 to calculate the pH.

**Calculate the pH of a solution of 0.0025 M H**_{2}SO_{4}.

_{2}SO

_{4}.

**Step 1:**

Because there are 2 Hs at the beginning of the formula of this acid, it is diprotic. Therefore, [H^{+}] = two times the concentration of the acid.

[H^{+}] = 2 x (0.0025 M)

[H^{+}] = 0.0050 M or 5.0 x 10^{-3} M

**Step 2:**

pH = -log [H^{+}] = -log [5.0 x 10^{-3}]

**Step 3:**

pH= -(-2.3)

pH = 2.3

## How to calculate pH when given the molarity of a base:

**Step 1:**

Determine if the base contributes 1 or 2 hydroxide ions to the solution. Bases that contribute 1 hydroxide ion are called monobasic. Bases that contribute 2 hydroxide ions are called dibasic. This will help you determine the concentration of the OH^{–}, which will, in turn, help us to calculate the [H^{+}] needed to calculate pH. If the base contributes one OH^{–} ion, the concentration of the OH^{–} is the same as the concentration of the base.

Example: 1.3 M NaOH has a concentration of OH^{–} = 1.3 M

If the base is dibasic, the concentration of the OH^{–} is two times the concentration of the base.

Example: 1.3 M Ca(OH)_{2} has a concentration of OH^{–} = 2 x (1.3 M) = 2.6 M

Once you have determined the concentration of OH^{–} (also denoted as [OH^{–}]), you can move to step 2.

At this point, there are two different paths that you can choose to solve your problem. Both are correct, and each works equally well. The first path was demonstrated in the video. The second path takes advantage of a formula you have not yet been introduced to.

**Path 1:**

Step 2:

Use the [OH^{–}] and K_{w} to solve for [H^{+}].

[H^{+}] [OH^{–}] = K_{w} = 1.0 x 10^{-14}

[H^{+}] = 1.0 x 10^{-14} / [OH^{–}]

Step 3:

Using the formula for pH (pH = -log [H^{+}]), plug in your value for [H^{+}].

Step 4:

Using your scientific calculator, calculate the value for pH.

**Path 2:**

It is important to note here that not all acids and bases completely ionize or dissociate in water. Those that do are called strong acids and strong bases. Notice the use of the word strong here. Often times, students think that the term strong refers to the pH of the acid or base. That is not true. Strong refers to how much the acid or base dissociates in water. The above methods for working pH problems will only work for strong acids and strong bases. The acid and base must 100% dissociate so that the concentration of the H^{+}or OH^{–} is equal to (or a multiple of) the concentration of the original acid or base. Acids and bases that do not ionize or dissociate completely are called weak acids and bases. To calculate their pH, you need to know something called the dissociation constant (K_{a} or K_{b}) for that acid or base. We will not work problems involving weak acids and bases in this class. So, it is safe to assume that the above methods work for all of the problems that you will encounter in this class.

Step 2: Using the formula for pOH (pOH = -log [OH^{–}]), plug in your value for [OH^{–}].

pOH = -log [OH^{–}]

Step 3:

Using your scientific calculator, calculate the value for pOH.

Step 4:Using the formula pH + pOH = 14.0, solve for pH. This formula is derived from the relationship [H^{+}] [OH^{–}] = K_{w} = 1.0 x 10^{-14}. You do not need to be worried with how it was derived, but you may find it useful.

pH = 14.0 – pOH

### Calculate the pH of a 0.0010 M NaOH solution.

**Step 1:**

NaOH has only one hydroxide ion in its formula. Therefore, the [OH^{–}] = the concentration of the base.

[OH^{–}] = 0.0010 M or 1.0 x 10^{-3} M

**Path 1:**

Step 2:

Use the [OH^{–}] and K_{w} to solve for [H^{+}].

[H^{+}] [OH^{–}] = K_{w} = 1.0 x 10^{-14}

[H^{+}] = 1.0 x 10^{-14} / [OH^{–}]

[H^{+}] = 1.0 x 10^{-14} / 1.0 x 10^{-3} M

[H+] = 1.0 x 10^{-11} M

Step 3:

pH = -log [H^{+}] = -log [1.0 x 10^{-11}]

Step 4: Using your scientific calculator, calculate the value for pH.

pH = -log [1.0 x 10^{-11}] = -(-11)

pH = 11

**Path 2:**

Step 2:

pOH = -log [OH-] = -log 1.0 x 10^{-3}

Step 3:

pOH = -log 1.0 x 10^{-3} = -(-3)

pOH = 3.0

Step 4:

pH = 14.0 – pOH

pH = 14.0 – 3.0

pH = 11

## How to calculate the molarity of either the H^{+} or OH^{–} when given the pH of a solution:

**Step 1:**

Using the formula for pH (pH = -log [H^{+}]), plug in your value for pH.

Divide both sides of the equation by the negative sign.

-pH = log [H^{+}]

**Step 2:**

Using your scientific calculator, calculate the value for pH.

This is done by taking the inverse log of the negative pH value.

### Human blood has a pH of 7.4. Calculate both the the [H^{+}] and [OH^{–}] concentration.

**Step 1:**

pH = -log [H^{+}]

7.4 = -log [H^{+}]

-7.4 = log [H^{+}]

**Step 2:**

-7.4 = log [H^{+}]

Type into your calculator the inverse log. You may do this by typing the 2nd function key, then the LOG key. Or, your calculator may have a key marked 10^{x}. This is equivalent to the inverse log function. Then, type in -7.4.

[H^{+}] = 4.0 x 10^{-8} M

In this problem, there is an additional step. You must use [H^{+}] [OH^{–}] = 1.0 x 10^{-14} to calculate the concentration of OH^{–}.

[H^{+}] [OH^{–}] = 1.0 x 10^{-14}

[4.0 x 10^{-8}] [OH^{–}] = 1.0 x 10^{-14}

[OH^{–}] = 2.5 x 10^{-7} M

## Self-Assessment and Practice

What is the pH of a 2.5 x 10^{-6} M solution of HCl?

## Answer

pH = -log [H^{+}] = -log [2.5 x 10^{-6}] = -(-5.6) = 5.6

What is the pH of a 0.020M Sr(OH)_{2} solution?

## Answer

[H^{+}][OH^{–}] = 1.0 x 10^{-14}

[H^{+}][2(2.0 x 10^{-2})] = 1.0 x 10^{-14}

[H^{+}] = 2.5 x 10^{-13}

pH = -log [H^{+}] = -log [2.5 x 10^{-13}] = -(-12.6) = 13

What is the pH of a 0.020M HCl solution?

## Answer

pH = -log [H^{+}] = =log [2.0 x 10^{-2}] = -(-1.7) = 1.7

Find the [H^{+}] and the [OH^{–}] of a solution with a pH of 3.494.

## Answer

pH = -log [H^{+}]

3.494 = -log [H^{+}]

[H^{+}] = 3.206 x 10^{-4} M

[H^{+}][OH^{–}] = 1.0 x 10^{-14}

[OH^{–}] = 3.1 x 10^{-11} M [

Is this solution acidic or basic?

## Answer

acidic