Problems #11 - 25

**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 Na_{2}CO_{3} 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 Na_{2}CO_{3}2HCl(aq) + Na

_{2}CO_{3}(aq) ---> 2NaCl(aq) + CO_{2}(g) + H_{2}O(ℓ)molar ratio of HCl to Na

_{2}CO_{3}is 2 to 1.0.0125 mol times 2 = 0.0250 mol of HCl was titrated.

2) The rest of the HCl titrated solid MgCO_{3}. How much HCl was involved?

total amount of HCl added ---> (1.00 mol/L) (0.145 L) = 0.145 molHCl that reacted with the MgCO

_{3}---> 0.145 mol minus 0.0250 mol = 0.120 mol

3) Stoichiometry:

2HCl(aq) + MgCO_{3}(s) ---> MgCl_{2}(aq) + CO_{2}(aq) + H_{2}O(ℓ)HCl to MgCO

_{3}molar ratio is 2 to 12 is to 1 as 0.120 mol is to x

x = 0.0600 mol of MgCO

_{3}reacted0.0600 mol times 84.313 g/mol = 5.06 g (to three sig figs)

**Problem #12:** A chemist measured the amount of CaCO_{3} 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 CO_{2} gas. The chemical reaction is this:

CaCO_{3}(s) + 2HCl (aq) ---> CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(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 CaCO_{3} in the antacid tablet?

**Solution:**

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

_{3}---> 0.0125 - 0.0024105 = 0.0100895 molFrom equation in question, 2 moles of HCl are required for every 1 mole of CaCO

_{3}reacted.moles CaCO

_{3}reacted ---> 0.0100895 mol divided by 2 = 0.00504475 molmass CaCO

_{3}present ---> 0.00504475 mol times 100.086 g/mol = 0.505 g (to three sig figs)mass percent CaCO

_{3}---> (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 KNO_{3}---> 5.00 g / 101.102 g/mol = 0.049455 molKNO

_{3}is produced thusly:HNO

_{3}(aq) + KOH(aq) ---> KNO_{3}(aq) + H_{2}O(ℓ)The one to one stoichiometry means that 0.049455 mol of HNO

_{3}reacted with 0.049455 mol of KOH to produce 0.049455 mol mol of KNO_{3}.

**Problem #14:** An organic acid has the formula HOOC(CH_{2})_{n}COOH. 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 molYou 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(CH

_{2})104 = (2) (12 + 32 + 1) + (n) (14)

104 = 90 + 14n

n = (104 - 90) / 14 = 1

The formula is HOOCCH

_{2}COOH

**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(CH

_{2})COOH

**Problem #15:** A 25.00 mL sample of HCl(aq) was added to a 0.1996 g sample of CaCO_{3}. All the CaCO_{3} reacted, leaving some excess HCl(aq).

CaCO_{3}(s) + 2HCl(aq) ---> CaCl_{2}(aq) + H_{2}O(ℓ) + CO_{2}(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) ---> BaCl_{2}(aq) + 2H_{2}O(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 molTwo 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 CaCO_{3}?

moles CaCO_{3}---> 0.1996 g / 100.086 g/mol = 0.001994285 molTwo mole HCl needed for every one mole of CaCO

_{3}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}---> CoX_{2}+ 2H_{2}OTwo 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 Lx = 0.2325 g

6) Molar mass of acid:

0.2325 g / 0.003623 mol = 64.2 g/molHNO

_{3}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:

H_{2}A + 2NaOH ---> Na_{2}A + 2H_{2}O1 : 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 NaOHHA + NaOH ---> NaA + H

_{2}OHA 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 + H_{2}O1:1 molar ratio

0.0927g / 0.00121992 mol = 76 g/mol

3) diprotic:

H_{2}A + 2NaOH ---> Na_{2}A + 2H_{2}O1 : 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 (C_{9}H_{8}O_{4}, 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 mLx = 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 gCaCO

_{3}+ 2HCl ---> CaCl_{2}+ CO_{2}+ H_{2}Ofor every one mole of CaCO

_{3}, we require two moles of HCl0.200 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 H_{2}SO_{4} that react:

2NaOH + H_{2}SO_{4}---> Na_{2}SO_{4}+ 2H_{2}OThe molar ratio between NaOH and H

_{2}SO_{4}is 2:1Therefore, 0.05120 moles of NaOH will neutralize 0.02560 moles of H

_{2}SO_{4}

4) Calculate molarity of H_{2}SO_{4} solution:

0.02560 mol / 0.02500 L = 1.024 M

**Problem #22:** Consider the reaction:

2HCl + Ba(OH)_{2}---> BaCl_{2}+ 2H_{2}O

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 2:1Therefore, 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}+ 3H_{2}SO_{4}---> Al_{2}(SO_{4})_{3}+ 6H_{2}OThe key thing is the molar ratio that Al(OH)

_{3}and H_{2}SO_{4}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 H_{2}SO_{4} needed:

2 is to 3 as 0.01923 mol is to xx = 0.028845 mol

4) Determine the volume of H_{2}SO_{4}:

0.028845 mol divided by 1.65 mol/L = 0.01748 LChanging 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 H_{2}SO_{4}?

**Solution:**

1) CaO reacts with water as follows:

CaO + H_{2}O ---> Ca(OH)_{2}

2) Calcium hydroxide racts with sulfuric acid as folows:

Ca(OH)_{2}+ H_{2}SO_{4}---> CaSO_{4}+ 2H_{2}O

3) Combining them gives:

CaO + H_{2}SO_{4}---> CaSO_{4}+ H_{2}OThe important point is that there is a 1:1 molar ratio between CaO and H

_{2}SO_{4}

4) Determine moles of H_{2}SO_{4}:

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

5) Determine grams of CaO required:

4.50 mol times 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 + H_{2}O1:1 (the HCl to NaOH ratio)

3) Determine volume of HCl required:

M_{1}V_{1}= M_{2}V_{2}(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:** In order to find the strength of a sample of H_{2}SO_{4}, 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% CaCO_{3} How many moles of it reacted?

7.00 g minus 2.20 g = 4.80 g4.80 g / 100.09 g/mol = 0.0479568 mol

2) The chemical reaction is as follows:

H_{2}SO_{4}+ CaCO_{3}---> CaSO_{4}+ CO_{2}+ H_{2}OThe key is the molar ratio between H

_{2}SO_{4}and CaCO_{3}. It is 1 to 1.

3) Determine moles, then grams of H_{2}SO_{4} reacted:

From the 1:1 molar ratio, we know that 0.0479568 mol of H_{2}SO_{4}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%