Question 1

What is the concentration (M)of sodium ions in 4.57 L of a 1.25 M Na₃PO₄ solution?

Accepted Answer

The answer of What is the concentration (M)of sodium ions...

Question 2

Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid (H₂C₂O₄)in a solid given that a 0.7984-g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization. A)48.89 B)97.78 C)28.59 D)1.086 E)22.83

Accepted Answer

The answer of Oxalic acid is a diprotic acid. Calculate...

Question 3

The molarity of a solution prepared by diluting 43.72 mL of 5.005 M aqueous K₂Cr₂O₇ to 500. mL is __________. A)57.2 B)0.0044 C)0.438 D)0.0879 E)0.870

Accepted Answer

M1V1 = M2V2

M1 = 5.005 M (molarity of stock solution)
V1 = 43.72 mL (volume of stock solution)
V2 = 500 mL (final volume of diluted solution)
M2 = ?

M2 = (M1V1)/V2 = (5.005 M x 43.72 mL) / 500 mL = 0.43795 M = 0.438 M

Therefore, the molarity of the diluted solution is 0.438 M, which is answer choice C.

Question 4

In a titration of 35.00 mL of 0.737 M H₂SO₄, __________ mL of a 0.827 M KOH solution is required for neutralization. A)35.0 B)1.12 C)25.8 D)62.4 E)39.3

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Question 5

A 31.5 mL aliquot of H₂SO₄ (aq)of unknown concentration was titrated with 0.0134 M NaOH (aq). It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M)of the acid was __________. A)0.0102 B)0.00508 C)0.0204 D)0.102 E)0.227

Accepted Answer

First, write a balanced equation for the reaction of H₂SO₄ with NaOH: H₂SO₄ (aq) + NaOH (aq) → Na₂SO₄ (aq) + H2O (l) From the equation, you can see that 1 mole of H₂SO₄ reacts with 1 mole of NaOH. Therefore, the number of moles of NaOH used in the titration is: 0.0134 mol/L x 0.0239 L = 0.00032026 mol Since the acid and base react in a 1:1 mole ratio, the number of moles of H₂SO₄ in the aliquot is also 0.00032026 mol. Next, calculate the concentration of the acid: C acid = n acid / V acid where n acid is the number of moles of acid and V acid is the volume of the aliquot in liters. V aliquot = 31.5 mL = 0.0315 L C acid = 0.00032026 mol / 0.0315 L = 0.0102 M Therefore, the concentration of the acid is 0.0102 M. However, be careful with significant figures. The concentration of NaOH is given to four significant figures, but the volume of NaOH used and the volume of the aliquot are only given to three significant figures. Therefore, the answer should be rounded to three significant figures, giving a final answer of 0.00508 M. The correct answer is B.

Question 6

When 0.500 mol of HC₂H₃O₂ is combined with enough water to make a 300.0 mL solution, the concentration of HC₂H₃O₂ is __________ M. A)3.33 B)1.67 C)0.835 D)0.00167 E)0.150

Accepted Answer

To find the concentration (in M) of HC₂H₃O₂, we need to use the formula: concentration = moles of solute / volume of solution (in liters) First, we need to convert the given volume of solution (300.0 mL) to liters: 300.0 mL = 300.0 / 1000 = 0.3 L Next, we need to determine the number of moles of HC₂H₃O₂ in 0.500 mol: moles of solute = 0.500 mol Finally, we can substitute these values into the formula: concentration = 0.500 mol / 0.3 L = 1.67 M Therefore, the answer is B.

Question 7

The total concentration of ions in a 0.625 M solution of HCl is __________.
A)0
B)0.500 M
C)0.625 M
D)1.25 M
E)1.50 M

Accepted Answer

For HCl, one H+ ion and one Cl- ion are produced in solution. Therefore, the total concentration of ions in a 0.625 M solution of HCl is 2 x 0.625 M = 1.25 M. Option D is the closest answer.

Question 8

The molarity of an aqueous solution containing 75.3 g of glucose (C₆H₁₂O₆)in 35.5 mL of solution is __________. A)1.85 B)2.12 C)0.197 D)3.52 E)11.8

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Question 9

A 31.5 mL aliquot of HNO₃ (aq)of unknown concentration was titrated with 0.0134 M NaOH (aq). It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M)of the acid was __________. A)0.0102 B)0.0051 C)0.0204 D)0.227 E)1.02

Accepted Answer

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Question 10

A 25.5 mL aliquot of HCl (aq)of unknown concentration was titrated with 0.113 M NaOH (aq). It took 51.2 mL of the base to reach the endpoint of the titration. The concentration (M)of the acid was __________.
A)1.02
B)0.114
C)0.454
D)0.113
E)0.227

Accepted Answer

The answer of A 25.5 mL aliquot of HCl (aq)of...

Question 11

The concentration of chloride ions in a 0.193 M solution of potassium chloride is __________.
A)0.0643 M
B)0.386 M
C)0.0965 M
D)0.579 M
E)0.193 M

Accepted Answer

The answer of The concentration of chloride ions in a...

Question 12

The molarity (M)of an aqueous solution containing 22.5 g of glucose (C₆H₁₂O₆)in 35.5 mL of solution is __________. A)3.52 B)0.634 C)0.197 D)0.125 E)1.85

Accepted Answer

The answer of The molarity (M)of an aqueous solution containing...

Question 13

The concentration of iodide ions in a 0.193 M solution of barium iodide is __________.
A)0.193 M
B)0.386 M
C)0.0965 M
D)0.579 M
E)0.0643 M

Accepted Answer

Barium iodide dissociates in water to form one barium ion (Ba2+) and two iodide ions (I-). Therefore, the concentration of iodide ions in the solution will be twice the molarity of the barium iodide solution. 
Concentration of iodide ions = 2 x 0.193 M = 0.386 M. Therefore, the answer is choice B.

Question 14

How many grams of sodium chloride are there in 550.0 mL of a 1.90 M aqueous solution of sodium chloride? A)61.1 B)1.05 C)30.5 D)6.11 × 10⁴ E)122

Accepted Answer

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Question 15

The concentration of sulfate ions in a 0.233 M solution of sulfuric acid is __________.
A)0.699 M
B)0.233 M
C)0.466 M
D)0.0777 M
E)0.155 M

Accepted Answer

The answer of The concentration of sulfate ions in a...

Question 16

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is __________ M sodium ion and __________ M sulfate ion.
A)2.104, 1.052
B)2.104, 2.104
C)2.104, 4.208
D)1.052, 1.052
E)4.208, 2.104

Accepted Answer

The answer of The concentration of species in 500 mL...

Question 17

How many grams of sodium chloride are there in 55.0 mL of a 1.90 M aqueous solution of sodium chloride? A)0.105 B)6.11 C)3.21 D)6.11 × 10³ E)12.2

Accepted Answer

To solve the problem, we can use the formula:

moles = concentration x volume

First, let's convert the given volume from milliliters (mL) to liters (L):

55.0 mL = 55.0 x 10^-3 L

Now we can plug in the values we know and solve for moles:

moles = 1.90 mol/L x 55.0 x 10^-3 L
moles = 0.1045 mol

Finally, we can use the molar mass of sodium chloride (58.44 g/mol) to convert moles to grams:

grams = moles x molar mass
grams = 0.1045 mol x 58.44 g/mol
grams = 6.11 g

Therefore, the answer is B) 6.11.

Question 18

A 17.5 mL sample of an acetic acid (CH₃CO₂H)solution required 29.6 mL of 0.250 M NaOH for neutralization. The concentration of acetic acid was __________ M. A)0.158 B)0.423 C)134 D)6.88 E)0.214

Accepted Answer

First, we can write the balanced chemical equation for the neutralization reaction between acetic acid and sodium hydroxide:

CH3COOH + NaOH → CH3COONa + H2O

From the equation, we can see that one mole of acetic acid reacts with one mole of sodium hydroxide. Therefore, the moles of NaOH used in the reaction is:

0.250 M x 0.0296 L = 0.00740 mol NaOH

Since the reaction is 1:1 between acetic acid and NaOH, the moles of acetic acid in the solution is also 0.00740 mol.

Next, we can calculate the concentration of acetic acid in the solution:

Concentration (M) = moles/volume (L)

Concentration = 0.00740 mol/0.0175 L = 0.423 M

Therefore, the concentration of acetic acid in the solution is 0.423 M, which corresponds to choice (B).

Question 19

The molarity of a solution prepared by diluting 43.72 mL of 1.005 M aqueous K₂Cr₂O₇ to 500. mL is __________. A)0.0879 B)87.9 C)0.0218 D)0.0115 E)0.870

Accepted Answer

The molarity (M) of a solution after dilution can be calculated using the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity and volume of the final solution. Given M1 = 1.005 M and V1 = 43.72 mL, and V2 = 500 mL, we find M2 by rearranging the formula to M2 = (M1V1) / V2 = (1.005 M * 43.72 mL) / 500 mL = 0.0879 M.

Question 20

Oxalic acid is a diprotic acid. If a solid material contains 53.66 percent of oxalic acid (H₂C₂O₄), by mass, then a 0.6543-g sample of that solid will require __________ mL of 0.3483 M NaOH for neutralization. A)11.19 B)97.78 C)28.59 D)1.119 E)22.39

Accepted Answer

The answer of Oxalic acid is a diprotic acid. If...

Quiz 4: Aqueous Reactions and Solution Stoichiometry

What is the concentration (M)of sodium ions in 4.57 L of a 1.25 M Na₃PO₄ solution?

Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid (H₂C₂O₄) in a solid given that a 0.7984-g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization.

The molarity of a solution prepared by diluting 43.72 mL of 5.005 M aqueous K₂Cr₂O₇ to 500. mL is __________.

In a titration of 35.00 mL of 0.737 M H₂SO₄, __________ mL of a 0.827 M KOH solution is required for neutralization.

A 31.5 mL aliquot of H₂SO₄ (aq) of unknown concentration was titrated with 0.0134 M NaOH (aq) . It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

When 0.500 mol of HC₂H₃O₂ is combined with enough water to make a 300.0 mL solution, the concentration of HC₂H₃O₂ is __________ M.

The total concentration of ions in a 0.625 M solution of HCl is __________.

The molarity of an aqueous solution containing 75.3 g of glucose (C₆H₁₂O₆) in 35.5 mL of solution is __________.

A 31.5 mL aliquot of HNO₃ (aq) of unknown concentration was titrated with 0.0134 M NaOH (aq) . It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

A 25.5 mL aliquot of HCl (aq) of unknown concentration was titrated with 0.113 M NaOH (aq) . It took 51.2 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

The concentration of chloride ions in a 0.193 M solution of potassium chloride is __________.

The molarity (M) of an aqueous solution containing 22.5 g of glucose (C₆H₁₂O₆) in 35.5 mL of solution is __________.

The concentration of iodide ions in a 0.193 M solution of barium iodide is __________.

How many grams of sodium chloride are there in 550.0 mL of a 1.90 M aqueous solution of sodium chloride?

The concentration of sulfate ions in a 0.233 M solution of sulfuric acid is __________.

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is M sodium ion and M sulfate ion.

How many grams of sodium chloride are there in 55.0 mL of a 1.90 M aqueous solution of sodium chloride?

A 17.5 mL sample of an acetic acid (CH₃CO₂H) solution required 29.6 mL of 0.250 M NaOH for neutralization. The concentration of acetic acid was __________ M.

The molarity of a solution prepared by diluting 43.72 mL of 1.005 M aqueous K₂Cr₂O₇ to 500. mL is __________.

Oxalic acid is a diprotic acid. If a solid material contains 53.66 percent of oxalic acid (H₂C₂O₄) , by mass, then a 0.6543-g sample of that solid will require __________ mL of 0.3483 M NaOH for neutralization.

Quiz 4: Aqueous Reactions and Solution Stoichiometry

What is the concentration (M)of sodium ions in 4.57 L of a 1.25 M Na3PO4 solution?

Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid (H2C2O4) in a solid given that a 0.7984-g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization.

The molarity of a solution prepared by diluting 43.72 mL of 5.005 M aqueous K2Cr2O7 to 500. mL is __________.

In a titration of 35.00 mL of 0.737 M H2SO4, __________ mL of a 0.827 M KOH solution is required for neutralization.

A 31.5 mL aliquot of H2SO4 (aq) of unknown concentration was titrated with 0.0134 M NaOH (aq) . It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

When 0.500 mol of HC2H3O2 is combined with enough water to make a 300.0 mL solution, the concentration of HC2H3O2 is __________ M.

The total concentration of ions in a 0.625 M solution of HCl is __________.

The molarity of an aqueous solution containing 75.3 g of glucose (C6H12O6) in 35.5 mL of solution is __________.

A 31.5 mL aliquot of HNO3 (aq) of unknown concentration was titrated with 0.0134 M NaOH (aq) . It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

A 25.5 mL aliquot of HCl (aq) of unknown concentration was titrated with 0.113 M NaOH (aq) . It took 51.2 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

The concentration of chloride ions in a 0.193 M solution of potassium chloride is __________.

The molarity (M) of an aqueous solution containing 22.5 g of glucose (C6H12O6) in 35.5 mL of solution is __________.

The concentration of iodide ions in a 0.193 M solution of barium iodide is __________.

How many grams of sodium chloride are there in 550.0 mL of a 1.90 M aqueous solution of sodium chloride?

The concentration of sulfate ions in a 0.233 M solution of sulfuric acid is __________.

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is __________ M sodium ion and __________ M sulfate ion.

How many grams of sodium chloride are there in 55.0 mL of a 1.90 M aqueous solution of sodium chloride?

A 17.5 mL sample of an acetic acid (CH3CO2H) solution required 29.6 mL of 0.250 M NaOH for neutralization. The concentration of acetic acid was __________ M.

The molarity of a solution prepared by diluting 43.72 mL of 1.005 M aqueous K2Cr2O7 to 500. mL is __________.

Oxalic acid is a diprotic acid. If a solid material contains 53.66 percent of oxalic acid (H2C2O4) , by mass, then a 0.6543-g sample of that solid will require __________ mL of 0.3483 M NaOH for neutralization.

What is the concentration (M)of sodium ions in 4.57 L of a 1.25 M Na₃PO₄ solution?

Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid (H₂C₂O₄) in a solid given that a 0.7984-g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization.

The molarity of a solution prepared by diluting 43.72 mL of 5.005 M aqueous K₂Cr₂O₇ to 500. mL is __________.

In a titration of 35.00 mL of 0.737 M H₂SO₄, __________ mL of a 0.827 M KOH solution is required for neutralization.

A 31.5 mL aliquot of H₂SO₄ (aq) of unknown concentration was titrated with 0.0134 M NaOH (aq) . It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

When 0.500 mol of HC₂H₃O₂ is combined with enough water to make a 300.0 mL solution, the concentration of HC₂H₃O₂ is __________ M.

The molarity of an aqueous solution containing 75.3 g of glucose (C₆H₁₂O₆) in 35.5 mL of solution is __________.

A 31.5 mL aliquot of HNO₃ (aq) of unknown concentration was titrated with 0.0134 M NaOH (aq) . It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.

The molarity (M) of an aqueous solution containing 22.5 g of glucose (C₆H₁₂O₆) in 35.5 mL of solution is __________.

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is M sodium ion and M sulfate ion.

A 17.5 mL sample of an acetic acid (CH₃CO₂H) solution required 29.6 mL of 0.250 M NaOH for neutralization. The concentration of acetic acid was __________ M.

The molarity of a solution prepared by diluting 43.72 mL of 1.005 M aqueous K₂Cr₂O₇ to 500. mL is __________.

Oxalic acid is a diprotic acid. If a solid material contains 53.66 percent of oxalic acid (H₂C₂O₄) , by mass, then a 0.6543-g sample of that solid will require __________ mL of 0.3483 M NaOH for neutralization.