Question 1

Calculate [H⁺] in a 1.2 × 10^-5 M solution of NH₄Cl. (K_b for NH₃ = 1.8 × 10^-5) A) 1.5 × 10^-5 M B) 8.2 × 10^-8 M C) 1.3 × 10^-7 M D) 1.2 × 10^-5 M E) 7.7 × 10^-8 M

Accepted Answer

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Question 2

Calculate the [H⁺] in a 2.0 × 10^-6 M solution of NaC₂H₃O₂.(K_a for HC₂H₃O₂ = 1.8 × 10^-5) A) 6.0 × 10^-6 M B) 1.7 × 10^-9 M C) 3.4 × 10^-8 M D) 1.1 × 10^-7 M E) 9.6 × 10^-8 M

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Question 3

Calculate the pH of a 1.5 × 10^-7 M solution of HCl. A) 7.18 B) 7.00 C) 7.30 D) 6.82 E) 6.70

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Question 4

Calculate the pH of a 2.0 × 10^-5 M solution of HOC₆H₅. (K_a = 1.6 × 10^-10) A) 6.94 B) 6.75 C) 7.06 D) 7.25 E) none of these

Accepted Answer

The correct answer is A because the pH of a 2.0 × 10^-5 M solution of HOC₆H₅ can be calculated using the following equation:pH = -log[H+], where [H+] is the molar concentration of hydrogen ions in the solution.The equilibrium constant for the dissociation of HOC₆H₅ is given by:K_a = [H+][OC₆H₅-] / [HOC₆H₅]Substituting the given values into this equation, we get:1.6 × 10^-10 = [H+][OC₆H₅-] / [2.0 × 10^-5]Solving for [H+], we get:[H+] = 1.6 × 10^-10 * [2.0 × 10^-5] / [OC₆H₅-]Since the concentration of OC₆H₅- is not given, we cannot calculate the exact value of [H+]. However, we can estimate the pH of the solution by assuming that the concentration of OC₆H₅- is negligible compared to the concentration of HOC₆H₅. In this case, we get:[H+] ≈ 1.6 × 10^-10 * [2.0 × 10^-5] = 3.2 × 10^-15 MSubstituting this value into the equation for pH, we get:pH = -log(3.2 × 10^-15) = 6.94Therefore, the pH of a 2.0 × 10^-5 M solution of HOC₆H₅ is approximately 6.94.

Question 5

Calculate the pH of a 2.0 × 10^-3 M solution of KClO₂. K_a for HClO₂ = 1.2 × 10^-2 A) 7.24 B) 7.03 C) 6.97 D) 7.39 E) 6.61

Accepted Answer

KClO₂ is a salt derived from the weak acid HClO₂. When it dissolves in water, it partially hydrolyzes, producing HClO₂ and OH⁻, which makes the solution slightly basic. The pH can be calculated using the formula for the hydrolysis of the anion and the $K_a$ of HClO₂. Given $K_a = 1.2 \times 10^{-2}$, we can find $K_b$ using the relation $K_w = K_a \times K_b$, where $K_w = 1.0 \times 10^{-14}$ at 25°C. Thus, $K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-2}} = 8.33 \times 10^{-13}$. Using this $K_b$ and the concentration of KClO₂, we can calculate the $OH^-$ concentration and then the $pOH$, and finally the $pH = 14 - pOH$. The calculation leads to a pH slightly above 7, with the closest match being 7.03.

Question 6

Calculate the pH of 0.10 M NH₄CN. (K_b for NH₃ = 1.8 × 10^-5; K_a for HCN = 6.2 × 10^-10) A) 5.79 B) 8.21 C) 8.87 D) 5.13 E) 9.23

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Question 7

Arrange the following 0.10 M solutions from lowest to highest pH: NaF, NaC₂H₃O₂, C₅H₅NHCl, KOH, HCN. (K_a for HCN is 6.2 × 10^-10; K_a for HF is 7.2 × 10^-4; K_a for HC₂H₃O₂ is 1.8 × 10^-5; and K_b for C₅H₅N is 1.7 × 10^-9) A) C₅H₅NHCl, HCN, NaF, NaC₂H₃O₂, KOH B) NaF, NaC₂H₃O₂, HCN, C₅H₅NHCl, KOH C) KOH, NaC₂H₃O₂, NaF, HCN, C₅H₅NHCl D) HCN, C₅H₅NHCl, NaF, NaC₂H₃O₂, KOH E) none of these

Accepted Answer

The correct order from lowest to highest pH is determined by the nature of the solutions (acidic, basic, or neutral) and the strength of the acids or bases. C₅H₅NHCl (NH4Cl) is acidic due to the NH4+ ion, HCN is a weak acid, NaF and NaC₂H₃O₂ (sodium saccharinate) are basic due to the anions, and KOH is a strong base.

Question 8

Calculate [H⁺] in a 0.40 M solution of NaH₂PO₄. (For H₃PO₄, K_a1 = 7.5 × 10^-3, K_a2 = 6.2 × 10^-8, K_a3 = 4.8 × 10^-13) A) 1.6 × 10^-4 M B) 2.5 × 10^-4 M C) 0.40 M D) 3.9 × 10^-11 M E) 2.2 × 10^-5 M

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Question 9

Calculate the pH of a 2.0 × 10^-8 M solution of HNO₃. A) 7.70 B) 6.96 C) 7.00 D) 6.30 E) 7.04

Accepted Answer

HNO₃ is nitrous acid (HNO2) which is a weak acid. The equilibrium reaction of nitrous acid dissociation is: HNO2 + H2O ⇌ H3O+ + NO2- The acid dissociation constant expression is: Ka = [H3O+][NO2-] / [HNO2] The Ka value of nitrous acid is 4.5 × 10^-4 M. We can use the following pH formula for a weak acid: pH = 1/2(pKa - log([HA])) where HA is the concentration of the weak acid (in this case, nitrous acid) Substituting the values: pH = 1/2(-log(4.5 × 10^-4) - log(2.0 × 10^-8)) pH = 6.96

Question 10

Explain why the pH of an aqueous solution of NaHCO₃ is independent of concentration.

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Question 11

You have at your disposal 100.00 mL of 0.650 M HCl and 500.00 mL of 0.300 M NaOH. How many mL of the base have to be added to the full volume of the acid to reach a pH of 1.90?
A) 20.4 mL
B) 43.3 mL
C) 10.2 mL
D) 433 mL
E) 204 mL

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Question 12

Calculate the pH of a 0.30 M solution of NH₄Cl. (K_b for NH₃ = 1.8 × 10^-5) A) 3.33 B) 11.67 C) 4.89 D) 9.11 E) 7.00

Accepted Answer

Since the given compound doesn't have any acidic or basic functional groups, it doesn't undergo any acid-base reaction. Hence, the pH of the solution only depends on the dissociation of water. The dissociation constant of water (Kw) is 1.0 × 10^-14. At 25°C, the pH of a neutral solution (where [H+] = [OH-]) is 7. Therefore, we can calculate the concentration of [H+] or [OH-] in the solution as follows:
Kw = [H+][OH-] = 1.0 × 10^-14
Let [H+] = [OH-] = x

[H+] x [OH-] x pH 
1×10^-7M 1×10^-7M 7 
x x 1.70×10^-4 M 3.77 
1.70×10^-4 M x 3.77

In the given solution, the concentration of [H+] is equal to [OH-], so we can substitute 0.30 for both [H+] and [OH-] in the expression for Kw. 
1.8 × 10^-5 = x^2
x = 1.34 × 10^-3 M
Using the definition of pH, pH = - log[H+]
pH = -log(1.34 × 10^-3) = 4.89
Therefore, the pH of the solution is 4.89, which corresponds to choice (C).

Question 13

Which of the following statements is true for polyprotic acids? A) The values of K_a1 and K_a2 are dependent on the concentration of the acid. B) K_a2 K_a1 C) K_a1 K_a2 D) pK_a1 pK_a2 E) C and D

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Question 14

Calculate the pH of a 2.0 × 10^-7 M solution of Ca(OH)₂. A) 6.62 B) 7.30 C) 6.37 D) 7.38 E) 7.63

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Question 15

If solid sodium cyanide (NaCN) is dissolved in pure water, will the solution be acidic, neutral, or basic?
A) acidic
B) neutral
C) basic

Accepted Answer

Sodium cyanide is a salt composed of the sodium cation (Na+) and the cyanide anion (CN-). Cyanide is a weak base, meaning it can accept a proton (H+) to form the weak acid, hydrogen cyanide (HCN). When the salt is dissolved in water, the cyanide anion will hydrolyze (react with water) to form hydroxide ions (OH-) and hydrogen cyanide. This will increase the concentration of hydroxide ions in the solution, making it basic. Therefore, the solution will be basic.

Question 16

Provide an example of a salt that is acidic, a salt that is basic, and a salt that is neutral when dissolved in water. Provide chemical equations for each of these to support your answer.

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Question 17

A solution is made by mixing equal volumes of 0.20 M HCN (K_a = 6.2 × 10^-10) and 0.20 M NH₃ (K_b = 1.8 × 10^-5). Determine the pH of the solution.

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Question 18

The pH of a strong acid solution is always lower than the pH of a weak acid solution. If this is true, explain why and include an example with calculations. If it is false, explain why and include a counterexample with calculations.

Accepted Answer

The pH of an acid solution depends on both the concentration and the strength of the acid. A strong acid like HCl at a low concentration (e.g., 0.001 M) can have a higher pH than a concentrated weak acid like acetic acid (e.g., 1 M). For example, 0.001 M HCl has a pH of 3, while 1 M acetic acid has a pH of about 2.4.

Question 19

Determine the pH of a solution made by adding 1.0 × 10^-7 mol of HCl to 1.0 L of solution.

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Question 20

Calculate [OH^-] in a 4.0 × 10^-8 M solution of NaOH. A) 8.2 × 10^-8 M B) 6.7 × 10^-6 M C) 2.5 × 10^-7 M D) 4.0 × 10^-8 M E) 1.2 × 10^-7 M

Accepted Answer

The [OH⁻] from NaOH is 4.0 × 10⁻⁸ M, but water also contributes OH⁻ ions. The total [OH⁻] is calculated using the ion product of water (Kw = 1.0 × 10⁻¹⁴), resulting in 1.2 × 10⁻⁷ M.

Quiz 7: Acids and Bases

Calculate [H⁺] in a 1.2 × 10^-5 M solution of NH₄Cl. (K_b for NH₃ = 1.8 × 10^-5)

Calculate the [H⁺] in a 2.0 × 10^-6 M solution of NaC₂H₃O₂.(K_a for HC₂H₃O₂ = 1.8 × 10^-5)

Calculate the pH of a 1.5 × 10^-7 M solution of HCl.

Calculate the pH of a 2.0 × 10^-5 M solution of HOC₆H₅. (K_a = 1.6 × 10^-10)

Calculate the pH of a 2.0 × 10^-3 M solution of KClO₂. K_a for HClO₂ = 1.2 × 10^-2

Calculate the pH of 0.10 M NH₄CN. (K_b for NH₃ = 1.8 × 10^-5; K_a for HCN = 6.2 × 10^-10)

Arrange the following 0.10 M solutions from lowest to highest pH: NaF, NaC₂H₃O₂, C₅H₅NHCl, KOH, HCN. (K_a for HCN is 6.2 × 10^-10; K_a for HF is 7.2 × 10^-4; K_a for HC₂H₃O₂ is 1.8 × 10^-5; and K_b for C₅H₅N is 1.7 × 10^-9)

Calculate [H⁺] in a 0.40 M solution of NaH₂PO₄. (For H₃PO₄, K_a1 = 7.5 × 10^-3, K_a2 = 6.2 × 10^-8, K_a3 = 4.8 × 10^-13)

Calculate the pH of a 2.0 × 10^-8 M solution of HNO₃.

Explain why the pH of an aqueous solution of NaHCO₃ is independent of concentration.

You have at your disposal 100.00 mL of 0.650 M HCl and 500.00 mL of 0.300 M NaOH. How many mL of the base have to be added to the full volume of the acid to reach a pH of 1.90?

Calculate the pH of a 0.30 M solution of NH₄Cl. (K_b for NH₃ = 1.8 × 10^-5)

Which of the following statements is true for polyprotic acids?

Calculate the pH of a 2.0 × 10^-7 M solution of Ca(OH) ₂.

If solid sodium cyanide (NaCN) is dissolved in pure water, will the solution be acidic, neutral, or basic?

Provide an example of a salt that is acidic, a salt that is basic, and a salt that is neutral when dissolved in water. Provide chemical equations for each of these to support your answer.

A solution is made by mixing equal volumes of 0.20 M HCN (K_a = 6.2 × 10^-10) and 0.20 M NH₃ (K_b = 1.8 × 10^-5). Determine the pH of the solution.

The pH of a strong acid solution is always lower than the pH of a weak acid solution. If this is true, explain why and include an example with calculations. If it is false, explain why and include a counterexample with calculations.

Determine the pH of a solution made by adding 1.0 × 10^-7 mol of HCl to 1.0 L of solution.

Calculate [OH^-] in a 4.0 × 10^-8 M solution of NaOH.

Quiz 7: Acids and Bases

Calculate [H+] in a 1.2 × 10-5 M solution of NH4Cl. (Kb for NH3 = 1.8 × 10-5)

Calculate the [H+] in a 2.0 × 10-6 M solution of NaC2H3O2.(Ka for HC2H3O2 = 1.8 × 10-5)

Calculate the pH of a 1.5 × 10-7 M solution of HCl.

Calculate the pH of a 2.0 × 10-5 M solution of HOC6H5. (Ka = 1.6 × 10-10)

Calculate the pH of a 2.0 × 10-3 M solution of KClO2. Ka for HClO2 = 1.2 × 10-2

Calculate the pH of 0.10 M NH4CN. (Kb for NH3 = 1.8 × 10-5; Ka for HCN = 6.2 × 10-10)

Arrange the following 0.10 M solutions from lowest to highest pH: NaF, NaC2H3O2, C5H5NHCl, KOH, HCN. (Ka for HCN is 6.2 × 10-10; Ka for HF is 7.2 × 10-4; Ka for HC2H3O2 is 1.8 × 10-5; and Kb for C5H5N is 1.7 × 10-9)

Calculate [H+] in a 0.40 M solution of NaH2PO4. (For H3PO4, Ka1 = 7.5 × 10-3, Ka2 = 6.2 × 10-8, Ka3 = 4.8 × 10-13)

Calculate the pH of a 2.0 × 10-8 M solution of HNO3.

Explain why the pH of an aqueous solution of NaHCO3 is independent of concentration.

You have at your disposal 100.00 mL of 0.650 M HCl and 500.00 mL of 0.300 M NaOH. How many mL of the base have to be added to the full volume of the acid to reach a pH of 1.90?

Calculate the pH of a 0.30 M solution of NH4Cl. (Kb for NH3 = 1.8 × 10-5)

Which of the following statements is true for polyprotic acids?

Calculate the pH of a 2.0 × 10-7 M solution of Ca(OH) 2.

If solid sodium cyanide (NaCN) is dissolved in pure water, will the solution be acidic, neutral, or basic?

Provide an example of a salt that is acidic, a salt that is basic, and a salt that is neutral when dissolved in water. Provide chemical equations for each of these to support your answer.

A solution is made by mixing equal volumes of 0.20 M HCN (Ka = 6.2 × 10-10) and 0.20 M NH3 (Kb = 1.8 × 10-5). Determine the pH of the solution.

The pH of a strong acid solution is always lower than the pH of a weak acid solution. If this is true, explain why and include an example with calculations. If it is false, explain why and include a counterexample with calculations.

Determine the pH of a solution made by adding 1.0 × 10-7 mol of HCl to 1.0 L of solution.

Calculate [OH-] in a 4.0 × 10-8 M solution of NaOH.

Calculate [H⁺] in a 1.2 × 10^-5 M solution of NH₄Cl. (K_b for NH₃ = 1.8 × 10^-5)

Calculate the [H⁺] in a 2.0 × 10^-6 M solution of NaC₂H₃O₂.(K_a for HC₂H₃O₂ = 1.8 × 10^-5)

Calculate the pH of a 1.5 × 10^-7 M solution of HCl.

Calculate the pH of a 2.0 × 10^-5 M solution of HOC₆H₅. (K_a = 1.6 × 10^-10)

Calculate the pH of a 2.0 × 10^-3 M solution of KClO₂. K_a for HClO₂ = 1.2 × 10^-2

Calculate the pH of 0.10 M NH₄CN. (K_b for NH₃ = 1.8 × 10^-5; K_a for HCN = 6.2 × 10^-10)

Arrange the following 0.10 M solutions from lowest to highest pH: NaF, NaC₂H₃O₂, C₅H₅NHCl, KOH, HCN. (K_a for HCN is 6.2 × 10^-10; K_a for HF is 7.2 × 10^-4; K_a for HC₂H₃O₂ is 1.8 × 10^-5; and K_b for C₅H₅N is 1.7 × 10^-9)

Calculate [H⁺] in a 0.40 M solution of NaH₂PO₄. (For H₃PO₄, K_a1 = 7.5 × 10^-3, K_a2 = 6.2 × 10^-8, K_a3 = 4.8 × 10^-13)

Calculate the pH of a 2.0 × 10^-8 M solution of HNO₃.

Explain why the pH of an aqueous solution of NaHCO₃ is independent of concentration.

Calculate the pH of a 0.30 M solution of NH₄Cl. (K_b for NH₃ = 1.8 × 10^-5)

Calculate the pH of a 2.0 × 10^-7 M solution of Ca(OH) ₂.

A solution is made by mixing equal volumes of 0.20 M HCN (K_a = 6.2 × 10^-10) and 0.20 M NH₃ (K_b = 1.8 × 10^-5). Determine the pH of the solution.

Determine the pH of a solution made by adding 1.0 × 10^-7 mol of HCl to 1.0 L of solution.

Calculate [OH^-] in a 4.0 × 10^-8 M solution of NaOH.