Ten Examples

The key point in the problems below will be the molar ratio between acid and base. Here are the ratios for the ten examples (acid first):

1) 1:1, 2) 1:1, 3) 1:2, 4) 1:2, 5) 1:1, 6) 1:3, 7) 1:2, 8) 2:3, 9) 1:1, 10) 1:1

There a bit of twist to the 1:1 ratios in 5, 9, and 10. In some of the solutions, I write the ratio the other way, as in writing 2:1 rather than 1:2. The order depends on the context specified.

A word to the wise: often 1:1 ratios are taught in class, but 2:1 ratios are tested. The 2:1 ratios may not show up in lecture, but will probably be included in your homework assignment.

There are two ways to solve the examples below: a step-by-step method or an all-in-one equation. The two are mathematically the same. I will use both ways in many problem and only one way in others. Check with your teacher as to the one you should use in class.

**Example #1:** If 20.60 mL of 0.0100 M aqueous HCl is required to titrate 30.00 mL of an aqueous solution of NaOH to the equivalence point, what is the molarity of the NaOH solution?

**Solution #1 (the step by step solution):**

1) Write the chemical equation for the reaction:

HCl + NaOH ---> NaCl + H_{2}O

2) The key molar ratio . . . :

. . . is that of HCl to NaOH, a 1 to 1 molar ratio.

3) Determine moles of HCl:

moles = MV = (0.0100 mol/L) (0.02060 L) = 0.000206 mol

4) Determine moles of NaOH:

1 is to 1 as 0.000206 mol is to xx = 0.000206 mol of NaOH consumed

5) Determine molarity of NaOH solution:

0.000206 mol / 0.03000 L = 0.00687 M

**Solution #2 (see a little below for a brief discussion on how the equation comes about):**

M _{a}V_{a}M _{b}V_{b}––––– = ––––– n _{a}n _{b}

(0.0100 mol/L) (20.60 mL) (x) (30.00 mL) –––––––––––––––––––––– = ––––––––––– 1 1 Another way you can see the solution written ---> [(0.0100 mol/L) (20.60 mL)] / 1 = [(x) (30.00 mL)] / 1

x = 0.00687 M

**Brief discussion about millimoles**

The example above can be solved via the concept of millimoles. This is the usual definition of a one molar solution:

1 mole / 1 L = 1 M

An alternate definition is this:

1000 millimoles / 1000 mL = 1 M

Therefore, this is true:

1 mmole / 1 mL = 1 MThe standard abbreviation for millimole is mmol.

Suppose we need to determine the moles of solute in 29.30 mL of 0.2080 M solution. We can, instead, determine the millimoles, as follows:

(0.2080 mmol / mL) (29.30 mL) = 6.0944 mmol

We can then proceed to calculate using millimoles rather than moles. Here is the step by step solution to #1 above:

mmoles = (0.0100 mmol/mL) (20.60 mL) = 0.2060 mmol1 is to 1 as 0.2060 mmol is to x

x = 0.2060 mmol of NaOH consumed

0.2060 mmol / 30.00 mL = 0.00687 M

In the examples and problems that follow, millimoles will show up in some problems, more or less at random.

**Example #2:** How many milliliters of 0.105 M HCl are needed to titrate 22.5 mL of 0.118 M NH_{3} to the equivalence point:

**Solution (using the step by step solution technique and moles):**

We will ignore the fact that HCl-NH_{3} is actually a strong-weak titration. We are only interested in the volume required for the equivalence point, not the pH at the equivalence point.

1) Chemical equation:

HCl + NH_{3}---> NH_{4}Cl

2) HCl to NH_{3} molar ratio:

1 : 1

3) Moles NH_{3}:

moles = MV = (0.118 mol/L) (0.0225 L) = 0.002655 mol

4) Determine moles of HCl used:

1 is to 1 as x is to 0.002655 molx = 0.002655 mol of HCl

5) Determine volume of HCl:

0.105 mol/L = 0.002655 mol / xx = 0.0253 L = 25.3 mL (to three sig figs)

**Solution (using the step by step solution technique and millimoles):**

(0.118 mmol/mL) (22.5 mL) = 2.655 mmol0.105 mmol/mL = 2.655 mmol / x

x = 25.3 mL (to three sig figs)

Does the molar ratio (remember, the ChemTeam maintains this is the key insight) always come from the coeffcients of the balanced chemical equation?

Yes.

Where did the second solution in Example #1 come from?

First, the final equation, which will work in ALL problems of the type being discussed:

M _{a}V_{a}M _{b}V_{b}––––– = ––––– n _{a}n _{b}

Sometimes seen like this:

M_{a}V_{a}/ n_{a}= M_{b}V_{b}/ n_{b}

The n_{a} stands for the balanced chemical equation's coefficient associated with the acid. The n_{b} stands for the coefficient of the base.

The above understanding of the meaning of n_{a} and n_{b} merits a bit of discussion.

These are true:

M_{a}V_{a}= n_{a}

M_{b}V_{b}= n_{b}Where n

_{a}and n_{b}are understood to be the moles in solution of the acid or the base.

Then, consider this:

1) For complete neutralization, let there be a 2:1 molar ratio required between acid and base (an example would be HCl and Ba(OH)_{2}).

2) That means that the moles of acid required are TWICE as much as the moles of base.

3) That means we have this be true:

M _{a}V_{a}n _{a}2 <--- coefficient of HCl in balanced chemical equation ––––– = ––––– = ––– M _{b}V_{b}n _{b}1 <--- coefficient of Ba(OH) _{2}in balanced chemical equation

Note that the above 2:1 ratio result because the M_{a}V_{a} is TWICE as big as the M_{b}V_{b} at neutralization.

Drop the 2/1, do a little cross-multiplying and dividing and you get this:

M _{a}V_{a}M _{b}V_{b}––––– = ––––– (where n _{a}and n_{b}are understood to be the coefficients of the balanced chemical equation)n _{a}n _{b}

The above equation always works, no matter what the ratio is.

I might add that many teachers use M_{a}V_{a} = M_{b}V_{b} for the 1:1 ratio problems without explaining that n_{a} and n_{b} both equal 1, so they drop out.

**Example #3:** 27.0 mL of 0.310 M NaOH is titrated with 0.740 M H_{2}SO_{4}. How many mL of H_{2}SO_{4} are needed to reach the end point?

**Solution #1:**

1) Millimoles NaOH present:

(0.310 mmol/mL) (27.0 mL) = 8.37 mmol

2) NaOH to H_{2}SO_{4} molar ratio is . . . :

. . . 2 : 1This can be seen from the balanced chemical equation:

2NaOH + H

_{2}SO_{4}---> Na_{2}SO_{4}+ 2H_{2}O

3) So:

2 is to 1 as 8.37 mmol is to x8.37 mmol divided by 2 = 4.185 mmol of H

_{2}SO_{4}required

4) Calculate volume of H_{2}SO_{4} required:

4.185 mmol divided by 0.740 mmol/mL = 5.66 mL (to three sig figs)

**Solution #2:**

M _{a}V_{a}M _{b}V_{b}––––– = ––––– n _{a}n _{b}

(0.740 mol/L) (x) (0.310 mol/L) (27.0 mL) ––––––––––––––– = ––––––––––––––––––– 1 2 (1) (0.310) (27.0) = (2) (0.740) (x)

x = 5.66 mL

**Example #4:** H_{2}SO_{4} reacts with NaOH, producing water and sodium sulfate. What volume of 2.00 M NaOH will be required to react completely with 75.0 mL of 0.500 M H_{2}SO_{4}?

**Solution:**

1) The chemical reaction:

H_{2}SO_{4}+ 2NaOH ---> Na_{2}SO_{4}+ 2H_{2}O

2) Moles H_{2}SO_{4}:

(0.500 mol/L) (0.0750 L) = 0.0375 mol

3) Two moles of NaOH are required to neutralize one mole of H_{2}SO_{4}.

(0.0375 mol) (2) = 0.0750 mol (of NaOH required)

4) Calculate the volume required:

moles divided by molarity0.0750 mol / 2.00 mol/L = 0.03750 L

0.03750 L = 37.5 mL (to three sig figs)

**Example #5:** How many milliliters of 0.116 M H_{2}SO_{4} will be needed to titrate 25.0 mL of 0.00840 Ba(OH)_{2} to the equivalence point:

**Solution:**

1) Chemical equation:

H_{2}SO_{4}+ Ba(OH)_{2}---> BaSO_{4}+ 2H_{2}O

2) Molar ratio:

1 : 1

3) M_{a}V_{a} / n_{a} = M_{b}V_{b} / n_{b}:

[(0.116) (x)] / 1 = [(0.00840) (25.0)] / 1x = 1.81 mL (to three sig figs)

The reason I wrote the example just above is because H_{2}SO_{4} and Ba(OH)_{2} show up often in problems where the ratio is not 1:1. I did not want you to gain the impression that chemicals such as H_{2}SO_{4} and Ba(OH)_{2} can NEVER be involved in a 1:1 ratio.

Another 1:1 ratio can be achieved by having Al(OH)_{3} and H_{3}PO_{4} react. Here's the chemical equation:

Al(OH)_{3}+ H_{3}PO_{4}---> AlPO_{4}+ 3H_{2}O

See Example #10 for more.

**Example #6:** A solution of 0.3094 M KOH is used to neutralize 19.50 mL of a H_{3}PO_{4} solution. If 28.93 mL of the KOH solution is required to reach the endpoint, what is the molarity of the H_{3}PO_{4} solution?

**Solution #1 (the step-by-step approach):**

1) Determine mllimoles of KOH used:

moles = (0.3094 mmol/ml) (28.93 mL) = 8.950942 mmol

2) The KOH : H_{3}PO_{4} is 3:1. Based on:

H_{3}PO_{4}(aq) + 3KOH(aq) ---> K_{3}PO_{4}(aq) + 3H_{2}O(ℓ)

3) Determine mmoles of H_{3}PO_{4} that got neutralized:

8.950942 mmol / 3 = 2.983647 mmol

4) Determine molarity of the phosphoric acid solution:

2.983647 mmol / 19.50 mL = 0.1530 M

**Solution #2 (one-step equation):**

M_{a}V_{a}/ n_{a}= M_{b}V_{b}/ n_{b}[(x) (19.50 mL)] / 1 = [(0.3094 M) (28.93 mL)] / 3

(3) (x) (19.50 mL) = (1) (0.3094 M) (28.93 mL)

x = 0.1530 M

**Example #7:** If 32.8 mL of a 0.162 M NaOH solution is required to titrate 25.0 mL of a solution of H_{2}SO_{4}, what is the molarity of the H_{2}SO_{4} solution?

**Solution with one equation:**

See below for balanced chemical equation.

M _{a}V_{a}M _{b}V_{b}––––– = ––––– n _{a}n _{b}

(x) (25.0 mL) (0.162 M) (32.8 mL) –––––––––– = –––––––––––––––– 1 2 (1) (0.162 M) (32.8 mL) = (2) (x) (25.0 mL)

x = 0.106 M

**Solution by steps:**

1) Write the chemical equation:

H_{2}SO_{4}+ 2OH ---> Na_{2}SO_{4}+ 2H_{2}OThe key will be the one-to-two molar ratio between sulfuric acid and sodium hydroxide.

2) Determine moles of NaOH:

(0.162 mol/L) (0.0328 L) = 0.0053136 mol

3) Use the molar ratio to determine moles of sulfuric acid consumed:

one is to two as x is to 0.0053136x = 0.0026568 mol

4) Determine the H_{2}SO_{4} molarity:

0.0026568 mol / 0.0250 L = 0.106 M (to three sig figs)

**Example #8:** What is the concentration of a Ca(OH)_{2} solution if 10.0 ml of 0.600 M H_{3}PO_{4} solution is required to completely neutralize 12.5 ml of the Ca(OH)_{2} solution?

**Solution:**

3Ca(OH)_{2}+ 2H_{3}PO_{4}---> Ca_{3}(PO_{4})_{2}+ 6H_{2}OThe key is to see the 3 : 2 molar ratio between Ca(OH)

_{2}and H_{3}PO_{4}.moles H

_{3}PO_{4}---> (0.600 mol/L) (0.0100 L) = 0.00600 mol3 is to 2 as x is to 0.00600 mol

x = 0.00900 mol of Ca(OH)

_{2}required0.00900 mol / 0.0125 L = 0.720 M

**Example #9:** A Ba(OH)_{2} solution has a molarity of 0.0850 M and is used to titrate 37.5 mL of 0.0550 M H_{2}S. What is the volume of barium hydroxide solution required for complete neutralization of the H_{2}S?

**Solution:**

H_{2}S(aq) + Ba(OH)_{2}(aq) ---> BaS(s) + 2H_{2}O(ℓ)

M _{a}V_{a}M _{b}V_{b}––––– = ––––– n _{a}n _{b}

(0.0550 mol/L) (37.5 mL) (0.0850 mol/L) (x) –––––––––––––––––––– = –––––––––––––– 1 1 (1) (0.0850 mol/L) (x) = (1) (0.0550 mol/L) (37.5 mL)

x = 24.3 mL

**Example #10:** An Al(OH)_{3} solution has a molarity of 0.0850 M and is used to titrate 37.5 mL of 0.0550 M H_{3}PO_{4}. What is the volume of aluminum hydroxide solution required for complete neutralization of the H_{3}PO_{4}?

**Solution:**

H_{3}PO_{4}+ Al(OH)_{3}---> AlPO_{4}+ 3H_{2}O

(0.0550 mol/L) (37.5 mL) (0.0850 mol/L) (x) –––––––––––––––––––– = –––––––––––––– 1 1 (1) (0.0850 mol/L) (x) = (1) (0.0550 mol/L) (37.5 mL)

x = 24.3 mL