Titration to the equivalence point using masses: Determine unknown molarity when a strong acid (base) is titrated with a strong base (acid)
Problems #1 - 10

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The five examples

Problems #11 - 20


Problem #1: How many mL of 1.80 M hydrochloric acid are required to completly react with 30.0 g of calcium hydroxide?

Solution:

1) moles of Ca(OH)2:

30.0 g / 74.0918 g/mol = 0.4049031 mol

2) chemical equation for the reaction:

2HCl + Ca(OH)2 ---> CaCl2 + 2H2O

The molar ratio between HCl and calcium hydroxide is 2:1

Two moles of HCl are used up for every one mole of Ca(OH)2 reacted.

3) Moles of HCl used:

0.4049031 mol times 2 = 0.8098062 mol

4) Volume of HCl required:

moles = MV

0.8098062 mol = (1.80 mol/L) (x)

x = 0.449892 L

x = 450. mL (to three sig figs, note the explicit decimal point)


Problem #2: How many milliliters of 0.122 M HCl would be required to titrate 6.45 g Ba(OH)2?

Solution:

1) The balanced equation for the reaction:

2HCl(aq) + Ba(OH)2(s) ---> BaCl2(aq) + 2H2O(ℓ)

2) Moles of Ba(OH)2:

6.45 g / 171.3438 g/mol = 0.0376436 mol

3) There is a 2 to 1 molar ratio between HCl and Ba(OH)2. Consequently:

0.0376436 mol times 2 = 0.0752872 mole of HCl used

4) Volume of HCl required:

0.0752872 mole / 0.122 mol/L = 0.617 L = 617 mL

Note that MV = mass / molar mass was not used. MV = mass / molar mass only works when there is a 1:1 molar ratio.


Problem #3: What is the molarity of your HCl solution if it take 35.25 mL to reach the second equivalence point of a titration against 0.2225 grams of standard sodium carbonate?

Solution #1:

Reaching the second equivalence point means this reaction:
2HCl(aq) + Na2CO3(s) ---> 2NaCl(aq) + CO2(g) + H2O(ℓ)

That's a 2:1 ratio: two moles of HCl are needed to react with one mole of sodium carbonate.

0.2225 g / 105.988 g/mol = 0.0020993 mol of Na2CO3

0.0020993 mol times 2 = 0.0041986 mol of HCl required

0.0041986 mol / 0.03525 L = 0.1191 M

Solution #2:

Consider this reaction:
HCl(aq) + Na2CO3(s) ---> NaHCO3(aq) + NaCl(aq)

That's the reaction for the first equivalence point.

35.25 mL divided by 2 to get to first eq. point ---> 17.625 mL

0.2225 g / 105.988 g/mol = 0.0020993 mol of sodium carbonate

HCl and Na2CO3 react in a 1 to 1 molar ratio to produce the NaHCO3

So, 0.0020993 mol of HCl was required.

0.0020993 mol / 0.017625 L = 0.1191 M


Problem #4: How many milliliters of 0.200 M HCl can react with 6.25 g CaCO3?

Solution:

CaCO3(s) + 2HCl(aq) ---> CaCl2(aq) + CO2(g) + H2O(ℓ)

moles CaCO3 ---> 6.25 g / 100.086 g/mol = 0.0624463 mol

From the balanced equation, 2 moles of HCl are needed for every mole of CaCO3.

moles HCl ---> 0.0624463 mol times 2 = 0.1248926 mol

volume HCl required ---> 0.1248926 divided by 0.200 mol/L = 0.624463 L

to three sig figs, 624 mL


Problem #5: What is the molarity of sulfuric acid if 43.90 mL of H2SO4 is required to neutralize 0.455 g of sodium hydrogen carbonate?

Solution:

1) The reaction is:

H2SO4(aq) + 2NaHCO3(aq) ---> Na2SO4(aq) + 2CO2(g) + 2H2O(ℓ)

The key ratio is that 1 mole of sulfuric acid is needed for every 2 moles of NaHCO3.

2) moles of NaHCO3:

0.455 g / 84.0059 g/mol = 0.0054163 mol

3) moles H2SO4 required:

0.0054163 mol / 2 = 0.00270815 mol

The fact that I divided flows out of how the problem was worded. Usually, it's worded so as to produce a multiplication. Not in this problem.

4) molarity:

0.00270815 mol / 0.04390 L = 0.0617 M

Problem #6: What is the molarity of nitric acid if 42.60 mL of HNO3 is required to neutralize 0.555 g of sodium carbonate?

Solution:

1) The reaction is:

2HNO3(aq) + Na2CO3(aq) ---> 2NaNO3(aq) + CO2 + H2O(ℓ)

The key molar ratio is that 2 moles of nitric acid are required for every one mole of sodium carbonate neutralized.

2) moles of Na2CO3:

0.555 g / 105.988 g/mol = 0.00523644 mol

3) moles nitric acid required:

0.00523644 mol times 2 = 0.01047288 mol

4) molarity:

0.01047288 mol / 0.04260 L = 0.246 M

Problem #7: What is the molarity of sodium hydroxide if 35.60 mL of NaOH is required to neutralize 0.631 g of oxalic acid, H2C2O4?

Solution #1:

1) The reaction is:

H2C2O4(aq) + 2NaOH(aq) ---> Na2C2O4(aq) + 2H2O(ℓ)

The key is the 1 to 2 molar ratio between oxalic acid and sodium hydroxide.

2) moles of oxalic acid:

0.631 g / 90.0338 g/mol = 0.0070085 mol

3) moles of NaOH required:

0.0070085 mol times 2 = 0.014017 mol

4) molarity:

0.014017 mol / 0.03560 L = 0.394 M

Solution #2:

This solution is based on the idea that it takes half the volume of sodium hydroxide solution to reach the first equivalence point.

1) The reaction to the first equivalence point is:

H2C2O4(aq) + NaOH(aq) ---> NaHC2O4(aq) + H2O(ℓ)

The key idea is that the molar ratio between the reactants is a one to one ratio.

2) moles of oxalic acid:

0.631 g / 90.0338 g/mol = 0.0070085 mol

3) moles of NaOH required:

0.0070085 mol (based on 1:1 ratio)

4) molarity:

0.0070085 mol / 0.0178 L = 0.394 M

Problem #8: What volume of 0.189 M barium hydroxide is required to neutralize 1.228 g of potassium hydrogen phthalate, KHC8H4O4 (204.23 g/mol)?

Solution:

1) The reaction is:

2KHC8H4O4(aq) + Ba(OH)2(aq) ---> BaC8H4O4(s) + K2C8H4O4(aq) + 2H2O(ℓ)

The key ratio is the 2 to 1 molar ratio between KHP and barium hydroxide.

2) moles KHP:

1.228 g / 204.23 g/mol = 0.006013 mol

3) moles barium hydroxide:

0.006013 mol / 2 = 0.0030065 mol

4) volume required:

0.189 mol/L = 0.0030065 mol / x

x = 0.0159 L (or 15.9 mL)


Problem #9: What volume of 0.765 M H3PO4 is required to exactly neutralize 2.000 g of calcium hydroxide?

Solution:

2.000 g / 74.0918 g/mol = 0.02699354 mol of Ca(OH)2

2H3PO4(aq) + 3Ca(OH)2(s) ---> Ca3(PO4)2(s) + 6H2O(ℓ)

Two moles of H3PO4 are required for every three moles of Ca(OH)2

2 is to 3 as x is to 0.02699354 mol

x = 0.01799569 mol of H3PO4 required

0.01799569 mol divided by 0.765 mol/L = 0.0235 L = 23.5 mL


Problem #10: A 2.61 M solution of phosphoric acid (H3PO4) is to be reacted with aluminum hydroxide to make magnesium phosphate and water. How many mL of the phosphoric acid solution are needed to react with 26.7 g of aluminum hydroxide?

Solution:

H3PO4(aq) + Al(OH)3(s) ---> AlPO4(s) + 3H2O(ℓ)

One mole of H3PO4 is required for every one mole of Al(OH)3 3

Due to 1 to 1 molar ratio between the reactants, 0.342292 mol of H3PO4 reacted.

0.342292 mol / 2.61 mol/L = 0.131 L = 131 mL (to three sig figs)


The five examples

Problems #11 - 20

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