Titration to the equivalence point using masses: Determine unknown molarity when a strong acid (base) is titrated with a strong base (acid)
Problems #11 - 20

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The five examples

Problems #1 - 10


Problem #11: Some pure magnesium carbonate was added to 145. mL of 1.00 M HCl. When the reaction had finished, the solution was acidic. 25.0 mL of 0.500 M Na2CO3 solution was required to neutralize the excess acid. What mass of magnesium carbonate was originally used?

Solution:

1) How much excess acid was present?

(0.500 mol/L) (0.0250 L) = 0.0125 mol of Na2CO3

2HCl(aq) + Na2CO3(aq) ---> 2NaCl(aq) + CO2(g) + H2O(ℓ)

molar ratio of HCl to Na2CO3 is 2 to 1.

0.0125 mol times 2 = 0.0250 mol of HCl was titrated.

2) The rest of the HCl titrated solid MgCO3. How much HCl was involved?

total amount of HCl added ---> (1.00 mol/L) (0.145 L) = 0.145 mol

HCl that reacted with the MgCO3 ---> 0.145 mol minus 0.0250 mol = 0.120 mol

3) Stoichiometry:

2HCl(aq) + MgCO3(s) ---> MgCl2(aq) + CO2(aq) + H2O(ℓ)

HCl to MgCO3 molar ratio is 2 to 1

2 is to 1 as 0.120 mol is to x

x = 0.0600 mol of MgCO3 reacted

0.0600 mol times 84.313 g/mol = 5.06 g (to three sig figs)


Problem #12: A chemist measured the amount of CaCO3 present in an antacid tablet. A tablet weighing 0.743 g was dissolved in 25.0 mL of 0.500 M HCl, and boiled to remove the CO2 gas. The chemical reaction is this:

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

The amount of HCl added was more than enough to react with all of the calcium carbonate present in the tablet

The excess HCl was back-titrated with 0.1500 M NaOH, requiring 16.07 mL of the NaOH solution to reach the endpoint.

What is the mass percent of CaCO3 in the antacid tablet?

Solution:

moles HCl added ---> (0.500 mol/L) (0.0250 L) = 0.0125 mol

moles HCl unreacted ---> (0.1500 mol/L) (0.01607 L) = 0.0024105

Note that moles NaOH used = moles HCl used because of 1 to 1 molar ratio between NaOH and HCl when they react.

moles HCl reacted with CaCO3 ---> 0.0125 - 0.0024105 = 0.0100895 mol

From equation in question, 2 moles of HCl are required for every 1 mole of CaCO3 reacted.

moles CaCO3 reacted ---> 0.0100895 mol divided by 2 = 0.00504475 mol

mass CaCO3 present ---> 0.00504475 mol times 100.086 g/mol = 0.505 g (to three sig figs)

mass percent CaCO3 ---> (0.505 g / 0.743 g) * 100 = 68.0% ( to three sig figs)


Problem #13: In a neutralization reaction 5.00 g of potassium nitrate are produced after complete consumption of the acid and base reactants. How many moles or acid and base were used?

Solution:

moles KNO3 ---> 5.00 g / 101.102 g/mol = 0.049455 mol

KNO3 is produced thusly:

HNO3(aq) + KOH(aq) ---> KNO3(aq) + H2O(ℓ)

The one to one stoichiometry means that 0.049455 mol of HNO3 reacted with 0.049455 mol of KOH to produce 0.049455 mol mol of KNO3.


Problem #14: An organic acid has the formula HOOC(CH2)nCOOH. 5.20 g of this acid requires 40.0 mL of 2.50 M sodium hydroxide for complete neutralization. What is the value of n in the formula of this acid ?

Solution #1:

Moles NaOH consumed ---> (2.50 mol/L) (0.0400 L) = 0.100 mol

You need two moles of NaOH in order to neutralize one mol of the acid.

moles acid neutralized ---> 0.100 mol divided by 2 = 0.0500 mol

The molar mass of the acid ---> 5.20 g / 0.0500 mol = 104 g/mol

The molar mass of the acid may be expressed this way:

104 = 2 MM(COOH) + n MM(CH2)

104 = (2) (12 + 32 + 1) + (n) (14)

104 = 90 + 14n

n = (104 - 90) / 14 = 1

The formula is HOOCCH2COOH

Solution #2:

Comment: the person whose solution is below tends to give extremely condensed explanations to questions on Yahoo Answers. This is an example of one. All I did was format it.

n(H+) => 2n(OH-)

0.05 = 5.2/(90 + 14n)

4.5 + 0.7n = 5.2

n = 1

thus:

HOOC(CH2)COOH


Problem #15: A 25.00 mL sample of HCl(aq) was added to a 0.1996 g sample of CaCO3. All the CaCO3 reacted, leaving some excess HCl(aq).

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

The excess HCl(aq) required 48.96 mL of a 0.01044 M barium hydroxide solution for complete neutralization.

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

What was the molarity of the original HCl(aq)?

Solution:

1) How much HCl was left over?

moles Ba(OH)2 ---> (0.01044 mol/L) (0.04896 L) = 0.0005111424 mol

Two moles HCl needed for every one mole of Ba(OH)2

moles HCl ---> 0.0005111424 mol times 2 = 0.0010222848 mol

2) How many moles of HCl needed to react with the CaCO3?

moles CaCO3 ---> 0.1996 g / 100.086 g/mol = 0.001994285 mol

Two mole HCl needed for every one mole of CaCO3

0.001994285 mol times 2 = 0.00398857 mol

3) Total moles HCl:

0.00398857 mol + 0.0010222848 mol = 0.0050108548 mol

4) Molarity of HCl:

0.0050108548 mol / 0.02500 L = 0.2004 M (to four sig figs)

Problem #16: 4.65 g of Co(OH)2 is dissolved in 500.0 mL of solution. 3.64 g of an unknown acid is dissolved in 250.0 mL of solution. 18.115 mL of the base is used to titrate 25.0 mL of the acid to its endpoint

a) Calculate the concentration of the base solution.
b) Calculate the molar mass of the acid and identify it.

Solution:

1) Molarity of base:

MV = grams / molar mass

(x) (0.5 L) = 4.65 g / 92.9468 g/mol

x = 0.100 mol/L

2) Moles of base used:

(0.100 mol/L) (0.018115 L) = 0.0018115 mol

3) We must now assume that the base is monoprotic because the next step is to determine the moles of acid that reacted. With that assumption, we have this:

2HX + Co(OH)2 ---> CoX2 + 2H2O

Two HX are used up for every Co(OH)2 reacted.

4) Moles of acid:

0.0018115 mol times 2 = 0.003623 mol

5) Grams of acid in the 0.025 L:

4.65 g is to 0.5000 L as x is to 0.0250 L

x = 0.2325 g

6) Molar mass of acid:

0.2325 g / 0.003623 mol = 64.2 g/mol

HNO3 weighs 63 g/mol


Problem #17: A 0.3017 g sample of a diprotic acid (molar mass = 126.07 g/mol) was dissolved in water and titrated with a 37.26 mL sample of sodium hydroxide. A 24.05 mL sample of this sodium hydroxide was then used to react with 0.2506 g of an unknown acid, which has been determined to be monoprotic. What is the molar mass of the unknown acid?

Solution:

1) moles of diprotic acid:

0.3017 g / 126.07 g/mol = 0.002393115 mol

2) moles NaOH required:

H2A + 2NaOH ---> Na2A + 2H2O

1 : 2 molar ratio

0.002393115 mol acid times 2 = 0.004786230 mol base

Note that we used the 1:2 ratio to determine moles of base from moles of acid.

3) molarity of NaOH solution:

0.004786230 mol / 0.03726 L = 0.128455 M (I won't round off too much yet.)

4) molar mass of monoprotic acid:

(0.128455 mol/L) (0.02405 L) = 0.00308934275 mol NaOH

HA + NaOH ---> NaA + H2O

HA and NaOH react in a 1:1 molar ratio

0.00308934275 mol HA reacted

0.2506 g / 0.00308934275 mol = 81.1 g/mol


Problem #18: 11.96 mL of 0.102 M NaOH was used to titrate a 0.0927 g sample of unknown acid to the endpoint using phenolphthalein as an indicator. What is the molecular weight of the acid if it is monoprotic? If it's diprotic?

Solution:

1) moles NaOH:

(0.102 mol/L) (0.01196 L) = 0.00121992 mol

2) monoprotic:

HA + NaOH ---> NaA + H2O

1:1 molar ratio

0.0927g / 0.00121992 mol = 76 g/mol

3) diprotic:

H2A + 2NaOH ---> Na2A + 2H2O

1 : 2 molar ratio

0.00121992 mol base / 2 = 0.00060996 mol acid

0.0927 g / 0.00060996 mol = 152 g/mol

Note that we used the 1:2 ratio to determine moles of acid from moles of base.


Problem #19: How many grams of Aspirin (C9H8O4, a monoprotic acid) are required to react with exactly 29.4 mL of a 0.2400% w/w solution of NaOH?

Solution::

1) Assume density of NaOH to be 1.00 g/mL

0.2400% w/w means 0.2400 g of NaOH per 100.0 g of solution. By the density, we know that our 100.0 g of solution occupies 100.0 mL

2) How much NaOH is in the 29.4 mL of solution?

0.2400 g over 100 mL equals x over 29.4 mL

x = 0.07056 g of NaOH

3) How many moles is this?

0.07056 g / 40.0 g/mol = 0.001764 mol

4) How many mole of aspirin reacts?

Due to the 1:1 molar ratio (monoprotic acid + NaOH), we know that 0.001764 mol of aspirin reacts.

0.001764 mol times 180.1582 g/mol = 0.318 g (to three sig figs)


Problem #20: We have 500. mg of a common stomach antiacid. How many mL of 0.100 M HCl can we neutralize? Assume the antiacid is 40% calcium carbonate by mass.

Solution:

500. mg * 0.40 = 200. mg = 0.200 g

CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O

for every one mole of CaCO3, we require two moles of HCl

0.200 g / 100.086 g/mol = 0.0019983 mol (keep a couple extra digits until the end)

0.0019983 mol x 2 = 0.0039966 mol <--- moles of HCl required

0.0039966 mol divided by 0.100 mol/L = 0.039966 L

Rounding off and putting it in mL leads to an answer of 40.0 mL (to three sig figs)


Bonus Problem: In order to find the strength of a sample of H2SO4, 100.0 g of it was taken and a piece of marble weighing 7.00 g placed in it. When the reaction ceased, the marble piece was removed, dried and was found to weigh 2.20 g. What is the percent strength of the sulfuric acid solution?

Solution:

1) We will assume the marble is 100% CaCO3 How many moles of it reacted?

7.00 g minus 2.20 g = 4.80 g

4.80 g / 100.09 g/mol = 0.0479568 mol

2) The chemical reaction is as follows:

H2SO4 + CaCO3 ---> CaSO4 + CO2 + H2O

The key is the molar ratio between H2SO4 and CaCO3. It is 1 to 1.

3) Determine moles, then grams of H2SO4 reacted:

From the 1:1 molar ratio, we know that 0.0479568 mol of H2SO4 reacted.

0.0479568 mol times 98.081 g/mol = 4.70365 g

4) Percent strength (in terms of mass/mass) of solution:

(4.70365 g / 100.0 g) times 100 = 4.70%

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The five examples

Problems #1 - 10