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

Problems #1-10

Problems #11-25

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The first five examples below use a 1:1 molar ratio to solve the problem. In the next five examples (that is, examples 6 to 10), there are some 2:1 ratios and a 3:2 ratio. The additional problems in the linked files contain 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.


Example #6: How many mL of 2.5 M H2SO4 are needed to neutralize 75.0 grams of NaOH?

Solution:

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

75.0 g / 40.0 g/mol = 1.875 mol

Based on the 1:2 molar ratio between H2SO4 and NaOH, we see that it takes two NaOH for every one H2SO4

1.875 mol / 2 = 0.9375 mol of H2SO4 neutralized

0.9375 mol / 2.5 mol/L = 0.375 L = 375 mL


Example #7: If 0.2501 grams of dry sodium carbonate requires 27.00 mL of HCl for complete reaction, what is the molar concentration of HCl?

Solution:

Na2CO3 + 2HCl ---> 2NaCl + CO2 + H2O

moles Na2CO3 ---> 0.2501 g / 105.988 g/mol = 0.0023597 mol

2 moles of HCl are required for every mole of Na2CO3

0.0023597 mol times 2 = 0.0047194 mol of HCl

0.0047194 mol / 0.02700 L = 0.1748 M


Example #8: A 250.0 cm3 solution of NaOH was prepared. 25.0 cm3 was used as a sample. The 25.0 cm3 solution required 28.2 cm3 of 0.100 mol dm-3 of HCl for neutralization. Calculate what mass of NaOH was dissolved to make up the original 250.0 cm3 solution.

Solution:

1) Important starting information:

HCl + NaOH ---> NaCl + H2O
Molar ratio between HCl and NaOH is 1 : 1

2) Moles HCl needed to neutralize the NaOH in the sample:

(0.100 mol dm-3) (0.0282 dm3) = 0.00282 mol

Note that the change from cm3 to dm3 is the same as the change from mL to L. There are 1000 cm3 in 1 dm3.

3) Moles of NaOH neutralize, then the NaOH molarity:

Based on the 1:1 molar ratio, we know that 0.00282 mol of HCl neutralized 0.00282 mol of NaOH.

0.00282 mol / 0.0250 dm3 = 0.1128 M

We could have used M1V1 = M2V2 to get the molarity of the NaOH. I chose to do the longer calculation.

4) Mass of NaOH is the 250.0 cm3 solution:

MV = mass / molar mass

(0.1128 mol dm-3) (0.2500 dm3) = x / 40.0 g/mol

x = 1.128 g (rounded to three sig figs, it is 1.13 g)


Example #9: What mass of calcium carbonate, CaCO3 will a 31.27 mL sample of 0.2442 M phosphoric acid, H3PO4, neutralize according to the reaction shown below? The molar mass of calcium carbonate is 100.09 g/mol.

Solution:

1) The reaction of interest is this:

3CaCO3(s) + 2H3PO4(aq) ---> Ca3(PO4)2(s) + 3CO2(g) + 3H2O(ℓ)

2) Moles of phosphoric acid that react:

moles = MV

moles = (0.2442 mol/L) (0.03127 L) = 0.007636134 mol

3) Determine moles of CaCO3 consumed:

The molar ratio between CaCO3 and H3PO4 is 3:2

3 is to 2 as x is to 0.007636134 mol

x = 0.011454201 mol of CaCO3 consumed

4) Determine mass of CaCO3 consumed:

(0.011454201 mol) (100.09 g/mol) = 1.146 g (to 4 sig figs)

Example #10: 1.314 g of sodium carbonate was dissolved and made up to a final volume of 250.0 mL. This solution was used to determine the concentration of a solution of hydrochloric acid. Four 25.0 mL samples of the acid were titrated with the sodium carbonate solution. The average titration volume required to reach the end point was 23.45 mL. Calculate the concentration of the hydrochloric acid solution.

Solution:

1) Determine the molarity of the sodium carbonate solution:

MV = mass / molar mass

(M) (0.2500 L) = 1.314 g / 105.99 g/mol

molarity = 0.04958958 M

2) Determine the moles of sodium carbonate in the average volume of 23.45 mL:

MV = moles

(0.04958958 mol/L) (0.02345 L) = moles

0.001162875651 mol

Note: be careful about which volume goes where. When I first solved this problem, I used 25.0 mL for the sodium carbonate. Oopsie!!

3) Write the chemical equation for the reaction:

Na2CO3 + 2HCl ---> 2NaCl + CO2 + H2O

The key point is the 1:2 molar ratio between sodium carbonate and HCl.

4) Determine moles of HCl consumed, then molarity of HCl solution:

(0.001162875651 mol) (2) = 0.002325751302 mol

0.002325751302 mol / 0.02500 L = 0.09303 M (to four sig figs)


Bonus Example #1: What is the formula weight of RbOH when 2.050 g is dissolved in 250.0 mL of water. 25.00 mL of the solution is neutralized by 18.18 mL of 0.1100 M HCl?

Solution:

RbOH + HCl ---> RbCl + H2O

The key point from the equation is that RbOH and HCl react in a 1:1 molar ratio.

Moles HCl that reacted ---> (0.1100 mol/L) (0.01818 L) = 0.0019998 mol

25.00 mL of the RbOH solution was neutralized by 0.0019998 mol of HCl. Therefore:

0.0019998 mol times 10 = 0.019998 mole of HCl needed for the entire 250.0 mL solution.

Based on the 1:1 molar ratio, we determine that 0.019998 mol of RbOH was in the 250.0 mL solution.

formula weight of RbOH ---> 2.050 g / 0.019998 mol = 102.5 g/mol


Bonus Example #2: A vinegar sample is titrated with 0.52 M NaOH. In one trial, 0.49 g of vinegar requires 0.73 g of NaOH to reach the endpoint.  Find the percent by mass of acetic acid in this vinegar sample. (Density of vinegar is 1.00 g/mL)

Solution:

1) That 0.73 g of NaOH means 0.73 g of NaOH solution. (Solving it with 0.73 g of NaOH gives an impossible answer.) In addition, we would need the density of the NaOH solution so as to determine the volume of solution. 0.52 M solution probably has a density close to 1.00 g/mL, so I'll use that.

0.73 g / 1.00 g/mL = 0.73 mL = 0.00073 L

0.52 mol/L times 0.00073 L = 0.0003796 mol of NaOH

This means 0.0003796 mol of acetic acid. (Based on the 1:1 molar ratio between NaOH and acetic acid when they react.)

0.0003796 mol times 60.0516 g/mol = 0.0227956 g

0.0227956 g / 0.49 g = 0.0465 = 4.6%

Note that the density of vinegar is not needed.

2) So, you're curious about that impossible answer?

0.73 g / 39.9969 g/mol = 0.0182514 mol of NaOH

0.0182514 mol of NaOH means 0.0182514 mol of acetic acid reacted.

0.0182514 mol times 60.0516 g/mol = 1.096 g of acetic acid

1.095 g / 0.49 g = 2.2367 = 223.67%


Problems #1-10

Problems #11-25

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