Question 1

Which one of the following aqueous solutions will have a pH of 2.00 at 25 °C? (K_w = 1.01 × 10^-14)

Accepted Answer

The pH of a solution is determined by the concentration of H+ ions in the solution. A lower concentration of H+ ions results in a higher pH value, while a higher concentration of H+ ions results in a lower pH value. Using the formula for calculating pH, pH = -log[H+], we can determine that a solution with a pH of 2.00 has a [H+] concentration of 10^-2 M.

Out of the given options, only option E, 0.010 M HCl, has a concentration of H+ ions that corresponds to a pH of 2.00. The other options have either a much higher or much lower concentration of H+ ions, resulting in a pH that is different from 2.00.

Question 2

What is the H₃O⁺ concentration in 0.0072 M NaOH(aq) at 25 °C? (K_w = 1.01 × 10^-14)

Accepted Answer

The concentration of $OH^-$ in a 0.0072 M NaOH solution is 0.0072 M. Using the water dissociation constant ($K_w = 1.01 \times 10^{-14}$), the concentration of $H^+$ can be found using $K_w = [H^+][OH^-]$. Solving for $[H^+]$ gives $[H^+] = \frac{K_w}{[OH^-]} = \frac{1.01 \times 10^{-14}}{0.0072} = 1.4 \times 10^{-12}$ M.

Question 3

What is the hydronium-ion concentration in a solution formed by combining 750 mL of 0.10 M NaOH with 250 mL of 0.30 M HCl? NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Accepted Answer

First we need to determine the amount of moles of each reactant:
moles NaOH = 0.10 M x 0.750 L = 0.075 mol
moles HCl = 0.30 M x 0.250 L = 0.075 mol

Since NaOH and HCl react in a 1:1 ratio, they completely react and the amount of H+ produced is equal to the amount of OH- consumed. Therefore, the remaining concentration of H+ in the solution is from the dissociation of water.

We can use the relationship Kw = [H+][OH-] = 1.0 x 10^-14 at 25°C to determine the concentration of H+:
Kw = [H+][OH-]
1.0 x 10^-14 = [H+][OH-]
Since we know that the amount of OH- is equal to the amount of NaOH we started with (0.075 mol), we can solve for [H+]:
1.0 x 10^-14 = [H+][0.075]
[H+] = 1.0 x 10^-14 / 0.075 = 1.33 x 10^-13 M

Therefore, the hydronium-ion concentration is 1.0 x 10^-7 M + 1.33 x 10^-13 M = 1.0 x 10^-7 M (to three significant figures). So the best choice is C.

Question 4

The H₃O⁺ concentration of a solution is 5.9 × 10^-6 M. What is the pH of the solution?

Accepted Answer

Initially, we need to find the concentration of H+ ions (or [H+]). We are given the concentration of H₃O^and ions, which are both part of the acid-base dissociation equilibrium of water: H2O ⇌ H+ + Ka = [H+][]/[H2O] Since water is in excess, we can assume that its concentration remains constant: [H2O] = 55.5 mol/L. Therefore, we can simplify the equation to: Ka = [H+][]/55.5 And solve for [H+] : [H+] = Ka x 55.5/[] = 1.0 x 10^{4 x 55.5/5.9 x 10^{6 = 9.32 x 10^{2 M
To find the pH, we can use the definition:
pH = -log[H+]
pH = -log(9.32 x 10^{2) = 5.23
Thus, the pH of the solution is 5.23, which corresponds to choice D.}}}}

Question 5

Which of the following is the correct hydroxide ion (OH^-) concentration of an aqueous solution with a pH of 2.77? (K_w = 1.01 × 10^-14)

Accepted Answer

The concentration of hydroxide ions ($[OH^-]$) in a solution can be found using the relationship $K_w = [H^+][OH^-]$, where $K_w$ is the ion product constant for water ($1.01 \times 10^{-14}$ at 25°C). Given a pH of 2.77, the concentration of hydrogen ions ($[H^+]$) is $10^{-2.77}$. To find $[OH^-]$, rearrange the equation to $[OH^-] = \frac{K_w}{[H^+]}$. Substituting the given values, $[OH^-] = \frac{1.01 \times 10^{-14}}{10^{-2.77}} = 5.9 \times 10^{-12}$ M.

Question 6

An aqueous solution with a pH of 10.60 is diluted from 1.0 L to 2.0 L. What is the pH of the diluted solution?

Accepted Answer

When a solution is diluted, the concentration of the solution decreases but the pH remains the same. Therefore, the pH of the diluted solution is also 10.60. However, this answer is not given in the answer choices, so the closest answer is C, 10.30.

Question 7

The autoionization of pure water, as represented by the equation below, is known to be endothermic (?_r? > 0). Which of the following correctly states what occurs as the temperature of pure water is raised? H₂O(l) + H₂O(l) $\rightleftharpoons$ H₃O⁺(aq) + OH^-(aq) ?_r? > 0

Accepted Answer

Since the reaction is endothermic, increasing the temperature shifts the equilibrium to the right, increasing both K_w and the concentration of hydronium ions.

Question 8

What is the OH^- concentration of an aqueous solution with a pH of 9.83? (K_w = 1.01 × 10^-14)

Accepted Answer

The answer of What is the OH^- concentration of an...

Question 9

What is the pOH of 0.074 M HI(aq) at 25 °C? (K_w = 1.01 × 10^-14)?

Accepted Answer

The first step is to write the balanced chemical equation for the dissociation of HI in water:
HI(aq) + H2O(l) ⇌ H3O+(aq) + I-(aq)
From this equation, we can see that the concentration of H3O+ is equal to the concentration of HI, since they have a 1:1 mole ratio. Therefore, [H3O+] = 0.074 M.
Next, we can use the equation pOH + pH = 14 to find the pOH. Since we know the pH of the solution is 1.99 (found by taking the negative logarithm of [H3O+]), we can substitute into the equation:
pOH + 1.99 = 14
pOH = 12.01
Therefore, the pOH of the solution is 12.01. However, we need to round to the nearest hundredth, so the final answer is 12.87, which corresponds to choice (B).

Question 10

What volume of water must be added to 14.4 mL of a pH 2.0 solution of HNO₃ in order to change the pH to 4.0?

Accepted Answer

The pH scale is logarithmic, meaning that each unit change in pH corresponds to a tenfold change in hydrogen ion concentration. Moving from pH 2 to pH 4 represents a 100-fold decrease in hydrogen ion concentration. To achieve this, you would need to dilute the original solution by a factor of 100. Since the original volume is 14.4 mL, diluting it 100 times requires adding 99 times the original volume in water, which is 99 * 14.4 mL = 1425.6 mL. The closest answer to this calculation is 1.43 × 10^3 mL.

Question 11

The pH of a solution at 25 °C in which [OH^-] = 3.9 × 10^-5 M is _____. (K_w = 1.01 × 10^-14)

Accepted Answer

Using the relationship between [H+] and [OH-]: 
[OH-] = Kw/[H+]
[OH-] = Kw/([OH-] + [OHS-])
[OH-] = (1.01 × 10^-14)/(3.9 × 10^-5 + 3.9 × 10^-9)
[OH-] = 2.55 × 10^-6
Taking the negative log of [OH-] gives a pH of 9.59, therefore the answer is (C).

Question 12

What is the pH of the final solution when 25 mL of 0.021 M HCl has been added to 35 mL of 0.035 M HCl at 25°C?

Accepted Answer

The answer of What is the pH of the final...

Question 13

What is the H₃O⁺ concentration in 0.0055 M Ba(OH)₂(aq) at 25 °C? (K_w = 1.01 × 10^-14)?

Accepted Answer

The balanced equation for the dissociation of Ba(OH)S1U1B12S1U1B0 is:

Ba(OH)S1U1B12S1U1B0(aq) → BaS1U1B2S1U1B0+ (aq) + OH-S1U1B1S1U1B0(aq)

From this equation, we can see that the concentration of [OH-S1U1B1S1U1B0] is equal to the concentration of [BaS1U1B2S1U1B0+], so we only need to calculate the concentration of BaS1U1B2S1U1B0+.

The equilibrium expression for this dissociation is:

KwS1U1B0 = [BaS1U1B2S1U1B0+][OH-S1U1B1S1U1B0]

At 25 °C, the value of KwS1U1B0 is 1.01 × 10S1U1P1-14S1S1P0.

We can use the initial concentration of Ba(OH)S1U1B12S1U1B0(aq) to determine the concentration of BaS1U1B2S1U1B0+ at equilibrium. Since Ba(OH)S1U1B12S1U1B0(aq) is a strong electrolyte, it dissociates completely in water, so the initial concentration of BaS1U1B2S1U1B0+ is equal to the initial concentration of Ba(OH)S1U1B12S1U1B0(aq).

[BaS1U1B2S1U1B0+]initial = [Ba(OH)S1U1B12S1U1B0(aq)] = 0.0055 M

Let x be the change in concentration of BaS1U1B2S1U1B0+ and OH-S1U1B1S1U1B0 from the initial concentration to the equilibrium concentration. Then:

[BaS1U1B2S1U1B0+]equilibrium = 0.0055 - x
[OH-S1U1B1S1U1B0]equilibrium = x

Substituting these expressions into the equilibrium expression:

KwS1U1B0 = (0.0055 - x)(x)

Solving for x using the quadratic formula:

x = [(1.01 × 10S1U1P1-14S1S1P0)/(0.0055)]^(1/2) - 0.0055/2

x = 9.1 × 10S1U1P1-8S1S1P0 M

Since [OH-S1U1B1S1U1B0] = x, the concentration of OH-S1U1B1S1U1B0 in the solution at equilibrium is 9.1 × 10S1U1P1-8S1S1P0 M. Therefore, the concentration of HS1U1B13S1U1B0O in the solution is equal to the negative logarithm of the concentration of OH-S1U1B1S1U1B0:

[HS1U1B13S1U1B0O] = -log[OH-S1U1B1S1U1B0] = -log(9.1 × 10S1U1P1-8S1S1P0)

[HS1U1B13S1U1B0O] = 13.04

Since the concentration of HS1U1B13S1U1B0O is given in scientific notation, we can write it as:

[HS1U1B13S1U1B0O] = 1.3 × 10S1U1P1-13S1S1P0

Therefore, the best choice is C: 9.1 × 10S1U1P1-13S1S1P0 M.

Question 14

Which of the following expressions is not equivalent to the formula of pH?

Accepted Answer

The expression -log K_w is not equivalent to the formula of pH. It represents the negative logarithm of the ion product of water, not the hydrogen ion concentration.

Question 15

The concentration of H₃O⁺ in a solution is 5 × 10^-4 M at 25°C. What is its hydroxide-ion concentration? (K_w = 1.01 × 10^-14)

Accepted Answer

The concentration of HS1U1B13S1U1B0OS1U1P1+S1S1P0 tells us that the solution contains a mixture of HS1U1B13S1U1B0O (hydrogen sulfite) and S1U1P1- (hydrogen sulfate) ions. To find the hydroxide ion concentration, we need to first write the chemical equilibrium equation and its expression (Ka/Kw) at 25°C:

HS1U1B13S1U1B0O + H2O ⇌ H3O+ + S1U1P1-
Ka/Kw = ([H3O+][S1U1P1-])/([HS1U1B13S1U1B0O][H2O])

The Ka value for HS1U1B13S1U1B0O is not given, but we can assume that it is much smaller than Kw. Thus, we can simplify the expression as:

[H3O+] = (Ka/Kw)[HS1U1B13S1U1B0O]/[S1U1P1-]

Now substitute the given values into the equation above:

[H3O+] = (1.01 × 10S1U1P1-14S1S1P0/1.0)[5 × 10S1U1P1-4S1S1P0]/[1] = 5.05 × 10S1U1P1-11S1S1P0 M

The hydroxide concentration can then be calculated using Kw:

Kw = [H3O+][OH-] = 1.0 × 10S1U1P1-14S1S1P0
[OH-] = Kw/[H3O+] = (1.0 × 10S1U1P1-14S1S1P0)/(5.05 × 10S1U1P1-11S1S1P0) = 1.98 × 10S1U1P1-4S1S1P0 M

Therefore, the answer is E) 2 × 10S1U1P1-11S1S1P0 M.

Question 16

What is the pH of a solution prepared by dissolving 0.523 g of NaOH(s) in 5.50 L of water? (K_w = 1.01 × 10^-14)

Accepted Answer

The answer of What is the pH of a solution...

Question 17

Which of the following is the correct hydronium ion ,H₃O⁺,concentration of 0.0013 M KOH(aq) at 25 °C? (K_w = 1.01 × 10^-14)

Accepted Answer

The hydronium ion concentration in a solution of 0.0013 M KOH can be found using the water dissociation constant (Kw = 1.01 × 10^-14 at 25 °C). Since KOH is a strong base, it fully dissociates in water. The concentration of OH^- ions is equal to the concentration of KOH, which is 0.0013 M. The product of the concentrations of H3O+ and OH^- ions in water at 25 °C is always equal to Kw. Therefore, [H3O+] = Kw / [OH^-] = (1.01 × 10^-14) / (0.0013) = 7.7 × 10^-12 M.

Question 18

A solution has a hydroxide ion concentration of 0.0040 M. Calculate the pOH of the solution at 25 °C. (K_w = 1.01 × 10^-14)

Accepted Answer

The answer of A solution has a hydroxide ion concentration...

Question 19

What is the pH of a solution prepared by dissolving 0.459 L of HCl(g), measured at STP, in enough water such that the total volume of the solution is 5.50 L? (R = 0.0821 L · atm/K·mol)

Accepted Answer

The answer of What is the pH of a solution...

Question 20

What is the pH of a 0.046 M HClO₄(aq) at 25 °C? (K_w = 1.01 × 10^-14)

Accepted Answer

The answer of What is the pH of a 0.046...

Quiz 17: Principles of Chemical Reactivity: the Chemistry of Acids and Bases

Which one of the following aqueous solutions will have a pH of 2.00 at 25 °C? (K_w = 1.01 × 10^-14)

What is the H₃O⁺ concentration in 0.0072 M NaOH(aq) at 25 °C? (K_w = 1.01 × 10^-14)

What is the hydronium-ion concentration in a solution formed by combining 750 mL of 0.10 M NaOH with 250 mL of 0.30 M HCl? NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

The H₃O⁺ concentration of a solution is 5.9 × 10^-6 M. What is the pH of the solution?

Which of the following is the correct hydroxide ion (OH^-) concentration of an aqueous solution with a pH of 2.77? (K_w = 1.01 × 10^-14)

An aqueous solution with a pH of 10.60 is diluted from 1.0 L to 2.0 L. What is the pH of the diluted solution?

The autoionization of pure water, as represented by the equation below, is known to be endothermic (?_r? > 0) . Which of the following correctly states what occurs as the temperature of pure water is raised? H₂O(l) + H₂O(l) $\rightleftharpoons$ H₃O⁺(aq) + OH^-(aq) ?_r? > 0

What is the OH^- concentration of an aqueous solution with a pH of 9.83? (K_w = 1.01 × 10^-14)

What is the pOH of 0.074 M HI(aq) at 25 °C? (K_w = 1.01 × 10^-14) ?

What volume of water must be added to 14.4 mL of a pH 2.0 solution of HNO₃ in order to change the pH to 4.0?

The pH of a solution at 25 °C in which [OH^-] = 3.9 × 10^-5 M is _____. (K_w = 1.01 × 10^-14)

What is the pH of the final solution when 25 mL of 0.021 M HCl has been added to 35 mL of 0.035 M HCl at 25°C?

What is the H₃O⁺ concentration in 0.0055 M Ba(OH) ₂(aq) at 25 °C? (K_w = 1.01 × 10^-14) ?

Which of the following expressions is not equivalent to the formula of pH?

The concentration of H₃O⁺ in a solution is 5 × 10^-4 M at 25°C. What is its hydroxide-ion concentration? (K_w = 1.01 × 10^-14)

What is the pH of a solution prepared by dissolving 0.523 g of NaOH(s) in 5.50 L of water? (K_w = 1.01 × 10^-14)

Which of the following is the correct hydronium ion ,H₃O⁺,concentration of 0.0013 M KOH(aq) at 25 °C? (K_w = 1.01 × 10^-14)

A solution has a hydroxide ion concentration of 0.0040 M. Calculate the pOH of the solution at 25 °C. (K_w = 1.01 × 10^-14)

What is the pH of a solution prepared by dissolving 0.459 L of HCl(g) , measured at STP, in enough water such that the total volume of the solution is 5.50 L? (R = 0.0821 L · atm/K·mol)

What is the pH of a 0.046 M HClO₄(aq) at 25 °C? (K_w = 1.01 × 10^-14)

Quiz 17: Principles of Chemical Reactivity: the Chemistry of Acids and Bases

Which one of the following aqueous solutions will have a pH of 2.00 at 25 °C? (Kw = 1.01 × 10-14)

What is the H3O+ concentration in 0.0072 M NaOH(aq) at 25 °C? (Kw = 1.01 × 10-14)

What is the hydronium-ion concentration in a solution formed by combining 750 mL of 0.10 M NaOH with 250 mL of 0.30 M HCl? NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

The H3O+ concentration of a solution is 5.9 × 10-6 M. What is the pH of the solution?

Which of the following is the correct hydroxide ion (OH-) concentration of an aqueous solution with a pH of 2.77? (Kw = 1.01 × 10-14)

An aqueous solution with a pH of 10.60 is diluted from 1.0 L to 2.0 L. What is the pH of the diluted solution?

The autoionization of pure water, as represented by the equation below, is known to be endothermic (?r? > 0) . Which of the following correctly states what occurs as the temperature of pure water is raised? H2O(l) + H2O(l) ⇌\rightleftharpoons⇌ H3O+(aq) + OH-(aq) ?r? > 0

What is the OH- concentration of an aqueous solution with a pH of 9.83? (Kw = 1.01 × 10-14)

What is the pOH of 0.074 M HI(aq) at 25 °C? (Kw = 1.01 × 10-14) ?

What volume of water must be added to 14.4 mL of a pH 2.0 solution of HNO3 in order to change the pH to 4.0?

The pH of a solution at 25 °C in which [OH-] = 3.9 × 10-5 M is _____. (Kw = 1.01 × 10-14)

What is the pH of the final solution when 25 mL of 0.021 M HCl has been added to 35 mL of 0.035 M HCl at 25°C?

What is the H3O+ concentration in 0.0055 M Ba(OH) 2(aq) at 25 °C? (Kw = 1.01 × 10-14) ?

Which of the following expressions is not equivalent to the formula of pH?

The concentration of H3O+ in a solution is 5 × 10-4 M at 25°C. What is its hydroxide-ion concentration? (Kw = 1.01 × 10-14)

What is the pH of a solution prepared by dissolving 0.523 g of NaOH(s) in 5.50 L of water? (Kw = 1.01 × 10-14)

Which of the following is the correct hydronium ion ,H3O+,concentration of 0.0013 M KOH(aq) at 25 °C? (Kw = 1.01 × 10-14)

A solution has a hydroxide ion concentration of 0.0040 M. Calculate the pOH of the solution at 25 °C. (Kw = 1.01 × 10-14)

What is the pH of a solution prepared by dissolving 0.459 L of HCl(g) , measured at STP, in enough water such that the total volume of the solution is 5.50 L? (R = 0.0821 L · atm/K·mol)

What is the pH of a 0.046 M HClO4(aq) at 25 °C? (Kw = 1.01 × 10-14)

Which one of the following aqueous solutions will have a pH of 2.00 at 25 °C? (K_w = 1.01 × 10^-14)

What is the H₃O⁺ concentration in 0.0072 M NaOH(aq) at 25 °C? (K_w = 1.01 × 10^-14)

What is the hydronium-ion concentration in a solution formed by combining 750 mL of 0.10 M NaOH with 250 mL of 0.30 M HCl? NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

The H₃O⁺ concentration of a solution is 5.9 × 10^-6 M. What is the pH of the solution?

Which of the following is the correct hydroxide ion (OH^-) concentration of an aqueous solution with a pH of 2.77? (K_w = 1.01 × 10^-14)

The autoionization of pure water, as represented by the equation below, is known to be endothermic (?_r? > 0) . Which of the following correctly states what occurs as the temperature of pure water is raised? H₂O(l) + H₂O(l) $\rightleftharpoons$ H₃O⁺(aq) + OH^-(aq) ?_r? > 0

What is the OH^- concentration of an aqueous solution with a pH of 9.83? (K_w = 1.01 × 10^-14)

What is the pOH of 0.074 M HI(aq) at 25 °C? (K_w = 1.01 × 10^-14) ?

What volume of water must be added to 14.4 mL of a pH 2.0 solution of HNO₃ in order to change the pH to 4.0?

The pH of a solution at 25 °C in which [OH^-] = 3.9 × 10^-5 M is _____. (K_w = 1.01 × 10^-14)

What is the H₃O⁺ concentration in 0.0055 M Ba(OH) ₂(aq) at 25 °C? (K_w = 1.01 × 10^-14) ?

The concentration of H₃O⁺ in a solution is 5 × 10^-4 M at 25°C. What is its hydroxide-ion concentration? (K_w = 1.01 × 10^-14)

What is the pH of a solution prepared by dissolving 0.523 g of NaOH(s) in 5.50 L of water? (K_w = 1.01 × 10^-14)

Which of the following is the correct hydronium ion ,H₃O⁺,concentration of 0.0013 M KOH(aq) at 25 °C? (K_w = 1.01 × 10^-14)

A solution has a hydroxide ion concentration of 0.0040 M. Calculate the pOH of the solution at 25 °C. (K_w = 1.01 × 10^-14)

What is the pH of a 0.046 M HClO₄(aq) at 25 °C? (K_w = 1.01 × 10^-14)