Examples #1 - 5

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The key point in the problems below will be the molar ratio between acid and base. Examples 1 and 2 utilize a 1:1 ratio and example 3 uses a 2:1 ratio. Example 4 is a 1:1 ratio (but with a twist) while Example 5 uses a 3:1 ratio.

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 be included in your homework assignment.

**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 general 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 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 (specific to a 1:1 ratio):**

M_{1}V_{1}= M_{2}V_{2}(0.0100 mol/L) (20.60 mL) = (x) (30.00 mL)

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

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 (abbreviated mmol) rather than moles. Here is the general solution to #1 above:

moles = MV = (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, 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 general 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 general 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.

Does the solution specific to the 1: 1 ratio ever work with other ratios?

No.

**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) 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)

**Example #4:** 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) Let's use the solution technique that only works with 1:1 ratios:

M_{1}V_{1}= M_{2}V_{2}(0.116 mol/L) (x) = (0.00840 mol/L) (25.0 mL)

x = 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.

**Example #5:** 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) 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

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