Question 1

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 pOH is calculated using the formula pOH = -log[OH⁻]. For a hydroxide ion concentration of 0.0040 M, pOH = -log(0.0040) = 2.40.

Question 2

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 constant.

pH = -log[H+]
10.60 = -log[H+]
[H+] = 2.511 x 10^-11 M

The new concentration of the solution after dilution is: 
2.511 x 10^-11 M / 2 = 1.256 x 10^-11 M

Therefore, the new pH of the diluted solution is:
pH = -log[H+]
pH = -log(1.256 x 10^-11)
pH = 10.30

Therefore, the answer is C.

Question 3

What is the hydroxide-ion concentration in a solution formed by combining 200.mL of 0.15 M HCl with 300.mL of 0.090 M NaOH at 25°C? HCl(aq)+ NaOH(aq)→ NaCl(aq)+ H₂O(l)

Accepted Answer

We need to use the balanced chemical equation to determine the amount of hydroxide ions produced from the reaction. From the equation, we see that the stoichiometry of the reaction is 1:1 between HCl and NaOH, so they will completely react with each other.

First, let's determine the amount of moles of HCl and NaOH used:

moles of HCl = (0.15 M) x (0.200 L) = 0.030 mol
moles of NaOH = (0.090 M) x (0.300 L) = 0.027 mol

Since HCl and NaOH react in a 1:1 ratio, the amount of HCl used is equal to the amount of NaOH used. Therefore, all of the HCl is neutralized by the NaOH, leaving only NaCl and water in solution.

Now, let's use the amount of moles of NaOH to calculate the concentration of hydroxide ions in the final solution:

moles of NaOH = moles of OH- ions produced = 0.027 mol
volume of final solution = 200 mL + 300 mL = 0.500 L

[OH-] = moles of OH- / volume of solution
[OH-] = 0.027 mol / 0.500 L
[OH-] = 0.054 M

Therefore, the hydroxide-ion concentration in the final solution is 0.054 M.

The answer is choice C.

Question 4

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 equation for the dissociation of HS1U1B13S1U1B0OS1U1P1+S1S1P0 is:

HS1U1B13S1U1B0OS1U1P1+S1S1P0 ⇌ H+ + S1U1B13S1U1B0O2- + S1U1P1- + S1S1P0

The equilibrium constant expression for this reaction is:

KS1U1B1wS1U1B0 = [H+][S1U1B13S1U1B0O2-][S1U1P1-]/[HS1U1B13S1U1B0OS1U1P1+S1S1P0]

We are given that the concentration of HS1U1B13S1U1B0OS1U1P1+S1S1P0 is 5 × 10S1U1P1-4S1S1P0 M. Let's assume that x is the concentration of H+ ions produced when HS1U1B13S1U1B0OS1U1P1+S1S1P0 dissociates.

Then, the concentration of S1U1B13S1U1B0O2- ions and S1U1P1- ions produced will also be x because the stoichiometric coefficients of all three ions are 1.

Substituting these values into the equilibrium constant expression and solving for x, we get:

1.01 × 10S1U1P1-14S1S1P0 = x2/(5 × 10S1U1P1-4S1S1P0)

x = 2 × 10S1U1P1-11S1S1P0 M, which is the concentration of H+ ions.

Now, we can use the equation for the ion product of water to calculate the hydroxide-ion concentration:

Kw = [H+][OH-]

1.01 × 10S1U1P1-14S1S1P0 = (2 × 10S1U1P1-11S1S1P0)([OH-])

[OH-] = 5 × 10S1U1P1-4S1S1P0 M

Therefore, the hydroxide-ion concentration is 5 × 10S1U1P1-4S1S1P0 M, which is option (A). However, this is not a listed option, and the closest listed option is (E), 2 × 10S1U1P1-11S1S1P0 M. This is because the answer key appears to have used a different value for Kw (potentially at a different temperature) than the one given in the question. This is a common issue with some chemistry problems, and it is important to understand the underlying concepts and equations in order to arrive at the correct answer even if the options do not match perfectly.

Question 5

Consider the reaction CO₃^2-(aq)+ H₂O(l) HCO₃^-(aq)+ OH^-(aq).K_b for CO₃^2- is 2.1 × 10^-4 at 25°C.What is K_a for the HCO₃^- ion at 25°C?

Accepted Answer

First, we can write the balanced chemical equation for the reaction: CO₃^2-(aq) + H₂O(l) ⇌ HCO₃^-(aq) + OH^-(aq) Using the given equilibrium constant: Kc = [HCO₃^-][OH^-] / [CO₃^2-][H₂O] Kc = 2.1 × 10^-4 We want to find the equilibrium constant, K_a, for the ion HCO₃^-. This ion is produced by the dissociation of the acid HCO₃^-: HCO₃^-(aq) ⇌ H+(aq) + CO₃^2-(aq) The equilibrium constant for this reaction is the acid dissociation constant, Ka: Ka = [H+(aq)][CO₃^2-(aq)] / [HCO₃^-(aq)] We can express [H+(aq)] in terms of [HCO₃^-(aq)] using the equilibrium expression for the acid dissociation reaction and the equilibrium constant for the overall reaction: Ka = [H+(aq)][CO₃^2-(aq)] / [HCO₃^-(aq)] Kc = [HCO₃^-(aq)][OH^-(aq)] / [CO₃^2-(aq)][H₂O(l)] [H+(aq)][CO₃^2-(aq)] / [HCO₃^-(aq)] = Kc / Ka [H+(aq)] = (Kc / Ka) x [H₂O(l)] / [OH^-(aq)] Substituting this expression for [H+(aq)] into the expression for Ka, we get: Ka = Kc x [H₂O(l)] / ([OH^-(aq)] x [HCO₃^-(aq)]) We can now substitute the given values of Kc and the concentrations into this equation to solve for Ka: Ka = (2.1 × 10^-4) x (1) / ([1] x [1]) = 2.1 × 10^-4 Finally, we can express K_a in terms of Ka using the expression: K_a = Kw / Ka where Kw is the ion product constant for water, which is 1.0 x 10^-14 at 25°C: K_a = (1.0 x 10^-14) / (2.1 x 10^-4) = 4.8 x 10^-11 Therefore, the answer is choice B, 4.8 x 10^-11.

Question 6

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 correct answer is A because the pH of a solution is determined by the concentration of hydrogen ions (H+). The concentration of H+ ions in a solution can be calculated using the following equation:[H+] = M * V / RTwhere:[H+] is the concentration of hydrogen ions in moles per literM is the molarity of the solution in moles per literV is the volume of the solution in litersR is the ideal gas constant (0.0821 L · atm/K·mol)T is the temperature in KelvinIn this problem, we are given the following information:V = 5.50 LT = 273 K (STP)R = 0.0821 L · atm/K·molM = 0.459 L / 5.50 L = 0.0835 MPlugging these values into the equation, we get:[H+] = 0.0835 M * 5.50 L / 0.0821 L · atm/K·mol * 273 K = 1.00 x 10^-2 MThe pH of a solution is defined as the negative logarithm of the hydrogen ion concentration:pH = -log[H+]Plugging in the value of [H+], we get:pH = -log(1.00 x 10^-2) = 2.429Therefore, the pH of the solution is 2.429.

Question 7

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 8

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 find out which reactant is limiting:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Using the balanced equation, we can see that the reaction consumes 1 mole of NaOH for every 1 mole of HCl, so we need to convert the volumes of each solution to moles:

(750 mL)(1 L/1000 mL)(0.10 mol/L) = 0.075 mol NaOH
(250 mL)(1 L/1000 mL)(0.30 mol/L) = 0.075 mol HCl

Since both reactants have the same number of moles, neither is in excess or limiting. Therefore, we need to use the stoichiometry of the balanced equation to find the concentration of hydronium ions (H3O+):

1 mol HCl produces 1 mol H3O+
0.075 mol HCl produces 0.075 mol H3O+
So, the concentration of H3O+ is:

[H3O+] = (0.075 mol)/(0.750 L + 0.250 L) = 0.0675 M

Since this is in scientific notation, we can write it as 6.75 × 10^-2 M or 1.0 × 10^-7 M after rounding to the appropriate number of significant figures. The answer choice closest to this is C, which is the correct answer.

Question 9

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 final concentration of HCl is calculated by combining the moles of HCl from both solutions and dividing by the total volume. The pH is then calculated using the formula pH = -log[H⁺].

Question 10

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 11

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 can be calculated using the equation pH = -log[H+]. A lower pH indicates a higher concentration of H+ ions. Among the given options, only option E, 0.010 M HCl, has a high enough concentration of H+ ions to have a pH of 2.00. Using the equation pH = -log[H+], we can find the concentration of H+ ions in the solution: pH = -log(0.010) = 2.00. Therefore, the correct answer is E.

Question 12

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 13

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 14

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

Accepted Answer

HClO is a strong acid, so it will dissociate completely in water to form H+ and ClO-. Therefore, the concentration of H+ in the solution will be equal to the concentration of HClO, which is 0.046 M. The pH of the solution can be calculated using the formula pH = -log[H+], which gives a pH of 1.34. Note that the value of the acid dissociation constant (K_w) for HClO is not needed to answer this question, as its status as a strong acid allows us to assume complete dissociation.

Question 15

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 16

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

Accepted Answer

The pOH can be calculated as 14 - pH = 4.17. The OH⁻ concentration is then 10^(-pOH) = 10^(-4.17) ≈ 6.8 × 10^(-5) M.

Question 17

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

To find the pH of the solution, we first calculate the pOH using the given hydroxide ion concentration ([OH⁻] = 3.9 × 10⁻⁵ M). The pOH is calculated as -log[OH⁻]. Then, we use the relationship pH + pOH = 14 (at 25 °C) to find the pH.pOH = -log(3.9 × 10⁻⁵) ≈ 4.41pH = 14 - pOH = 14 - 4.41 = 9.59Therefore, the correct answer is C, which corresponds to a pH of 9.59.

Question 18

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

Accepted Answer

The pOH of a solution can be found using the formula pOH = 14 - pH. Since HI is a strong acid, it dissociates completely in water, meaning the concentration of H+ ions is equal to the initial concentration of HI, which is 0.074 M. The pH is calculated using the formula pH = -log[H+], which gives pH = -log(0.074) ≈ 1.13. Therefore, pOH = 14 - 1.13 = 12.87.

Question 19

Hydrofluoric acid has a pK_a value of 3.14.What is the value of pK_b for fluoride ion?

Accepted Answer

The value of $pK_{a}$ for hydrofluoric acid (HF) is 3.14. The value of $pK_{b}$ for the fluoride ion (F-) can be found using the relationship $pK_{a} + pK_{b} = 14$ for water at 25°C. Therefore, $pK_{b} = 14 - 3.14 = 10.86$.

Question 20

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(aq) is:

Ba(OH)S1U1B12S1U1B0(aq) -> BaS1U1B2S1U1B0S1S1P0(aq) + 2OH-S1U1B1-(aq)

Kw = [H+S1U1B1(aq)][OH-S1U1B1(aq)] = 1.01 × 10S1U1P1-14S1S1P0

From the balanced equation, we see that the concentration of H+S1U1B1(aq) is twice the concentration of BaS1U1B2S1U1B0S1S1P0(aq), so:

[H+S1U1B1(aq)] = 2[BaS1U1B2S1U1B0S1S1P0(aq)]

Substituting this into the expression for Kw, we get:

Kw = 4[BaS1U1B2S1U1B0S1S1P0(aq)][OH-S1U1B1(aq)]

At equilibrium, the initial concentration of BaS1U1B2S1U1B0S1S1P0(aq) will decrease by an amount equal to the concentration of OH-S1U1B1(aq), and the concentration of H+S1U1B1(aq) will increase by the same amount. Let x be the concentration of OH-S1U1B1(aq) at equilibrium. Then:

[BaS1U1B2S1U1B0S1S1P0(aq)] = 0.0055 - x
[H+S1U1B1(aq)] = 2(0.0055 - x)

Substituting these expressions into the equation for Kw, we get:

1.01 × 10S1U1P1-14S1S1P0 = 4(0.0055 - x)(x)

Solving for x, we get:

x = 9.1 × 10S1U1P1-14S1S1P0

Therefore, the concentration of OH-S1U1B1(aq) at equilibrium is 9.1 × 10S1U1P1-14S1S1P0 M. Since the initial concentration of H+S1U1B1(aq) is negligible compared to the equilibrium concentration, we can assume that the concentration of H+S1U1B1(aq) is equal to 2(0.0055 - x) = 1.1 × 10S1U1P1-2S1S1P0 M. The concentration of BaS1U1B2S1U1B0S1S1P0(aq) at equilibrium is equal to 0.0055 - x = 0.0055 - 9.1 × 10S1U1P1-14S1S1P0 = 0.0055 M.

Therefore, the concentration of HS1U1B13S1U1B0OS1U1P1+S1S1P0 at equilibrium is equal to the concentration of H+S1U1B1(aq) plus the concentration of HS1U1B13S1U1B0OS1U1P1-S1U1B0 formed by the reaction of OH-S1U1B1(aq) with the initial concentration of HS1U1B13S1U1B0OS1U1P1+S1S1P0:

[HS1U1B13S1U1B0OS1U1P1+S1S1P0] = [H+S1U1B1(aq)] + [HS1U1B13S1U1B0OS1U1P1-S1U1B0]
[HS1U1B13S1U1B0OS1U1P1-S1U1B0] = x

Substituting the values of x and [H+S1U1B1(aq)], we get:

[HS1U1B13S1U1B0OS1U1P1+S1S1P0] = (1.1 × 10S1U1P1-2S1S1P0) + (9.1 × 10S1U1P1-14S1S1P0) = 1.0 × 10S1U1P1-13S1S1P0

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

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

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)

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?

What is the hydroxide-ion concentration in a solution formed by combining 200.mL of 0.15 M HCl with 300.mL of 0.090 M NaOH at 25°C? HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

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)

Consider the reaction CO₃^2-(aq) + H₂O(l) HCO₃^-(aq) + OH^-(aq) .K_b for CO₃^2- is 2.1 × 10^-4 at 25°C.What is K_a for the HCO₃^- ion at 25°C?

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)

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)

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)

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?

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

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)

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)

What is the pH of a 0.046 M HClO₄(aq) at 25 °C? (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)

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

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

Hydrofluoric acid has a pK_a value of 3.14.What is the value of pK_b for fluoride ion?

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

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

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)

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?

What is the hydroxide-ion concentration in a solution formed by combining 200.mL of 0.15 M HCl with 300.mL of 0.090 M NaOH at 25°C? HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

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)

Consider the reaction CO32-(aq) + H2O(l) HCO3-(aq) + OH-(aq) .Kb for CO32- is 2.1 × 10-4 at 25°C.What is Ka for the HCO3- ion at 25°C?

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)

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)

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)

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?

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

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)

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)

What is the pH of a 0.046 M HClO4(aq) at 25 °C? (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)

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

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

Hydrofluoric acid has a pKa value of 3.14.What is the value of pKb for fluoride ion?

What is the H3O+ concentration in 0.0055 M Ba(OH) 2(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.(K_w = 1.01 × 10^-14)

What is the hydroxide-ion concentration in a solution formed by combining 200.mL of 0.15 M HCl with 300.mL of 0.090 M NaOH at 25°C? HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

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)

Consider the reaction CO₃^2-(aq) + H₂O(l) HCO₃^-(aq) + OH^-(aq) .K_b for CO₃^2- is 2.1 × 10^-4 at 25°C.What is K_a for the HCO₃^- ion at 25°C?

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)

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 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)

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)

What is the pH of a 0.046 M HClO₄(aq) at 25 °C? (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)

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

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

Hydrofluoric acid has a pK_a value of 3.14.What is the value of pK_b for fluoride ion?

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