Titration to the equivalence point using masses: Determine unknown molarity when a strong acid (base) is titrated with a strong base (acid)
Problems #11 - 25

Ten examples

Problems #1-10

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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) (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) (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) (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 of 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: a rather condensed solution to the problem.

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) (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 moles HCl needed for every one mole of CaCO3

(0.001994285 mol) (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 Co(OH)2 (the base) is used to titrate 25.0 mL of the unknown 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) (0.5000 L) = 4.65 g / 92.9468 g/mol

x = 0.100057 mol/L

2) Moles of base used:

(0.100057 mol/L) (0.018115 L) = 0.00181253 mol

3) We must now assume that the acid 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:

(0.00181253 mol) (2) = 0.00362506 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 / 0.00362506 mol = 64.13687 g/mol

To three sig figs, this is 64.1 g/mol

HNO3 weighs 63.0 g/mol (to three sig figs)


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) (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) (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)


Problem #21: 32.00 g of sodium hydroxide were dissolved in 250.0 mL of solution to prepare the titrant. 25.00 mL of sulfuric acid were titrated with above titrant. It took 16.00 mL of sodium hydroxide solution to titrate to the end point. What is the molarity of the sulfuric acid?

Solution:

1) Determine molarity of NaOH:

MV = grams divided by molar mass

(x) (0.2500 L) = 32.00 g / 40.00 g/mol

x = 3.200 M

2) Determine moles NaOH in 16.00 mL of 3.200 M solution:

moles NaOH ⇒ (3.200 mol/L) (0.01600 L) = 0.05120 mol

3) Determine moles of H2SO4 that react:

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

The molar ratio between NaOH and H2SO4 is 2:1

Therefore, 0.05120 moles of NaOH will neutralize 0.02560 moles of H2SO4

4) Calculate molarity of H2SO4 solution:

0.02560 mol / 0.02500 L = 1.024 M

Problem #22: Consider the reaction:

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

How many mL of 0.7000 M HCl solution would just react with 8.000 g of Ba(OH)2?

Solution:

1) Calculate moles of Ba(OH)2:

8.000 g / 171.3438 g/mol = 0.04669 mol

2) Determine moles of HCl required to neutralize:

The molar ratio between HCl and Ba(OH)2 is 2:1

Therefore, 0.09338 moles of HCl is required to neutralize 0.04669 moles of Ba(OH)2

3) Determine volume of HCl required:

0.7000 mol/L = 0.09338 mol / x

x = 0.1334 L = 133.4 mL


Problem #23: How many mL of 1.65 M sulfuric acid will it take to fully react with 1.50 g of aluminum hydroxide?

Solution:

1) The chemical reaction is this:

2Al(OH)3 + 3H2SO4 ---> Al2(SO4)3 + 6H2O

The key thing is the molar ratio that Al(OH)3 and H2SO4 react in. It is a 2:3 ratio.

2) We need to know the moles of Al(OH)3:

1.50 g / 78.0027 g/mol = 0.01923 mol <--- I'll keep at least one guard digit to the end

3) Determine the moles of H2SO4 needed:

2 is to 3 as 0.01923 mol is to x

x = 0.028845 mol

4) Determine the volume of H2SO4:

0.028845 mol divided by 1.65 mol/L = 0.01748 L

Changing to mL and rounding off to three sig figs gives 17.5 mL for the final answer.


Problem #24: How much anhydrous CaO would be needed to neutralize 0.900 L of 5.00 M H2SO4?

Solution:

1) CaO reacts with water as follows:

CaO + H2O ---> Ca(OH)2

2) Calcium hydroxide racts with sulfuric acid as folows:

Ca(OH)2 + H2SO4 ---> CaSO4 + 2H2O

3) Combining them gives:

CaO + H2SO4 ---> CaSO4 + H2O

The important point is that there is a 1:1 molar ratio between CaO and H2SO4

4) Determine moles of H2SO4:

(0.900 L) (5.00 mol/L) = 4.50 mol

Because of the 1:1 molar ration between sulfuric acid and calcium oxide, the moles of sulfuric acid consumed equals the number of moles of CaO consumed.

5) Determine grams of CaO required:

(4.50 mol) (56.077 g/mol) = 252 g (to three sig fig)

Problem #25: How many milliliters of 0.105 M HCl are needed to neutralize 125.0 mL of a solution that contains 1.35 g of NaOH per liter.

Solution (using the solution technique specific to 1:1 ratios):

1) Determine molarity of NaOH solution:

MV = mass / molar mass

(x) (1.00 L) = 1.35 g / 40.00 g/mol

x = 0.03375 M (I'll round off the final answer to the problem.)

2) Write the chemical equation and state the key molar ratio:

HCl + NaOH ---> NaCl + H2O

1:1 (the HCl to NaOH ratio)

3) Determine volume of HCl required:

M1V1 = M2V2

(0.03375 mol/L) (125.0 mL) = (0.105 mol/L) (x)

x = 40.18 mL (to three sig figs, this is 40.2 mL)


Bonus Problem #1: 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) (98.081 g/mol) = 4.70365 g

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

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

Bonus Problem #2: Find the molar mass of Compound X (a monoprotic acid) from the following information:

Mass of X titrated: 0.1260 g
Molarity of NaOH used: 0.08810 M
Volume of NaOH solution used: 9.490 mL

However, there is one twist: your instructor INSISTS on the solution being done in Dimensional Analysis style.

Solution: the first few steps are a discussion leading up to the DA.

1) The equation to solve this problem is this:

  mass
MV =  –––––––––
  molar mass

2) Rearrange it to this:

  mass
molar mass =  ––––––
  MV

3) Then, flip it to get it aligned with DA style (the answer coming at the end of the calculations):

mass  
––––––  = molar mass
MV  

4) The mass must be divided by both the molarity AND the volume in order to get the molar mass. One important note: the volume must be in liters. The last conversion of the DA does that.

5) Here it is:

  1 L   1   1000 mL  
0.1260 g x  ––––––––––  x  ––––––––  x  –––––––  = 150.7 g/mol
  0.08810 mol   9.490 mL   1 L  

Ten examples

Problems #1-10

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