Titration to the equivalence point using masses: Determine unknown molarity when a strong acid (base) is titrated with a strong base (acid)
Examples #1 - 5

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Problems #1 - 10

Problems #11 - 20


All five of the problems below use a 1:1 molar ratio to solve the problem. The additional problems linked above contains a number of 2:1 molar ratio problems, several 1:1 ratios and one 3:2 ratio.


Example #1: How many milliliters of 0.122 M HCl would be required to titrate 6.45 g KOH?

Solution #1:

1) The balanced equation for the reaction:

HCl(aq) + KOH(s) ---> KCl(aq) + H2O(ℓ)

2) The HCl and the KOH react in a 1:1 molar ratio. This means:

moles HCl used = moles KOH used

3) Consequently, we can use:

MV = mass / molar mass

(0.122 mol/L) (x) = 6.45 g / 56.1049 g/mol

x = 0.9423215 L

to three sig figs, 942 mL

However, using the above only works when there is a 1:1 molar ratio between reactants. Below is the more general solution.

Solution #2:

1) moles KOH:

6.45 g / 56.1049 g/mol = 0.114963 mol

2) molar ratio:

From the balanced equation, we see that the HCl:KOH molar ratio is 1:1. Therefore:

0.114963 mol of HCl was used

3) volume of HCl:

0.114963 mol / 0.122 mol/L = 0.94232 L

to three sig figs, 942 mL


Example #2: What is the molarity of a hydrochloric acid solution if 25.0 mL of the solution reacts completely with 1.66 g NaHCO3?

Solution:

1) The balanced equation for the reaction:

HCl(aq) + NaHCO3(s) ---> NaCl(aq) + CO2(g) + H2O(ℓ)

2) Note the 1 to 1 molar relationship between HCl and NaHCO3.

moles NaHCO3 ---> 1.66 g / 84.0059 g/mol = 0.01976 mol

3) The 1:1 ratio means:

0.01976 mol of HCl reacted with 0.01976 mol of NaHCO3

0.01976 mol / 0.0250 L = 0.790 M (to three sig figs)


Example #3: In the following acid-base neutralization, 1.68 g of the solid acid phenol (HC6H5O; MW = 94.12 g/mol) neutralized 11.61 mL of aqueous NaOH solution. Calculate the molarity of the base solution.

Solution:

1) The chemical reaction is this:

HC6H5O(aq) + NaOH(aq) ---> NaC6H5O(aq) + H2O(ℓ)

The key is that there is a one-to-one molar ratio between the phenol (formula also seen as C6H5OH. However, it's an acid) and the sodium hydroxide.

2) Moles phenol:

1.68 g / 94.12 g/mol = 0.01785 mol

3) Molarity of NaOH solution:

Due to the one-to-one molar ratio, 0.01785 mol of NaOH reacted.

0.01785 mol / 0.01161 L = 1.54 M


Example #4: How many mL of 0.258 M NaOH are required to completely neutralize 2.00 g of acetic acid (HC2H3O2)?

Solution:

1) The chemical reaction is this:

HC2H3O2(aq) + NaOH(aq) ---> NaC2H3O2(aq) + H2O(ℓ)

The key is that there is a one-to-one molar ratio between the acetic acid and the sodium hydroxide.

2) Moles acetic acid:

2.00 g / 60.0516 g/mol = 0.0333047 mol

3) Volume of NaOH solution:

Due to the one-to-one molar ratio, 0.0333047 mol of NaOH reacted.

0.0333047 mol / 0.258 mol/L = 0.129 L = 129 mL

Comment: both phenol and acetic acid are weak acids. However, since we are not concerned with the pH of the solution, there is no need to distinguish between strong and weak. In other words, it takes exactly the same amount of base to neutralize a given amount of acid, either strong or weak.


Example #5: When 5.231 g of calcium hydroxide is reacted with a 29.0 mL of a 0.100 M sulfuric acid solution, what volume of the H2SO4 solution is required for complete neutralization?

Solution:

1) The chemical reaction is this:

H2SO4(aq) + Ca(OH)2(aq) ---> CaSO4(s) + 2H2O(ℓ)

The key is that there is a one-to-one molar ratio between the sulfuric acid and the calcium hydroxide.

2) Moles calcium hydroxide:

5.231 g / 74.0918 g/mol = 0.0706016 mol

3) Volume of H2SO4 solution:

Due to the one-to-one molar ratio, 0.0706016 mol of H2SO4 reacted.

0.0706016 mol / 0.100 mol/L = 0.706 L = 706 mL

Notice that both calcium hydroxide and sulfuric acid often show up in problems that use a 2:1 molar ratio. However, when paired together, a 1:1 ratio is used. The same can be said for a problem involving H3PO4 and Al(OH)3. A 1:1 molar ratio would be used. See problem #9 in the first file of additional problems.


Problems #1 - 10

Problems #11 - 20

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