Question 1

What is the molar solubility of AgCl in 0.10 M NH₃? K_sp for AgCl is 1.8 × 10^-10 and the K_f for Ag(NH₃)₂⁺ is 1.7 × 10⁷. A)1.3 × 10^-5 M B)5.0 × 10^-3 M C)5.5 × 10^-3 M D)5.5 × 10^-2 M

Accepted Answer

Using the information given, we can set up the following equilibrium expression: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] We know that the concentration of Cl- comes from the dissociation of NH₃: [NH₃] = [Cl-] We also know that Ag+ reacts with NH₃ to form a complex ion: Ag+(aq) + NH₃(aq) ⇌ Ag(NH₃)(aq) Kf = [Ag(NH₃)]/[Ag+][NH₃] Now we can use the equilibrium constant expressions to set up an expression for the molar solubility of AgCl: Ksp = [Ag+][Cl-] = [Ag+][NH₃] Kf = [Ag(NH₃)]/[Ag+][NH₃] Substituting [Cl-] = [NH₃] into the Ksp expression and solving for [Ag+], we get: [Ag+] = sqrt(Ksp/[Cl-]) = sqrt(1.8 × 10^-10/0.10) = 1.34 × 10^-5 M Substituting [Ag+] = 1.34 × 10^-5 M into the Kf expression and solving for [Ag(NH₃)], we get: [Ag(NH₃)] = Kf[Ag+][NH₃] = (1.7 × 10⁷)(1.34 × 10^-5)(0.10) = 2.41 × 10^-7 M Finally, we can use the stoichiometry of the reaction to find the molar solubility of AgCl: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) -1 +1 +1 If [Ag+] = 1.34 × 10^-5 M, then [Cl-] = [NH₃] = 1.34 × 10^-5 M And if [Ag(NH₃)] = 2.41 × 10^-7 M, then [AgCl] = [Ag+] = 1.34 × 10^-5 M Therefore, the molar solubility of AgCl is 1.34 × 10^-5 M, which corresponds to answer choice B.

Question 2

Sodium hypochlorite,NaOCl,is the active ingredient in household bleach.What is the concentration of hypochlorite ion if 20.00 mL of bleach requires 28.30 mL of 0.500 M HCl to reach the equivalence point?
A)0.208 M
B)0.353 M
C)0.708 M
D)1.21 M

Accepted Answer

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Question 3

Calculate the molar solubility of thallium(I)chloride in 0.30 M NaCl at 25°C.K_sp for TlCl is 1.7 × 10^-4. A)5.1 × 10^-5 M B)5.7 × 10^-4 M C)7.1 × 10^-3 M D)1.3 × 10^-2 M

Accepted Answer

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Question 4

The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid H₂X⁺)are K_a1 = 4.6 × 10^-3 and K_a2 = 2.0 × 10^-10.What is the pH of 50.00 mL of a 0.100 M solution of alanine after 100.00 mL of 0.100 M NaOH has been added? A)9.70 B)10.69 C)11.11 D)12.70

Accepted Answer

The correct answer is C because the pH of the solution is 11.11. The pKa of the first dissociation is 2.34 and the pKa of the second dissociation is 9.70. The addition of NaOH will cause the first dissociation to occur, resulting in the formation of H2X+ and OH-. The OH- will then react with H2X+ to form HX and H2O. The HX will then dissociate to form X- and H+. The pH of the solution will be determined by the concentration of H+ ions, which will be equal to the concentration of X- ions. The concentration of X- ions can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([X-]/[HX]). Substituting the values of pKa and [HX] into the equation, we get: pH = 9.70 + log([X-]/[HX]) = 9.70 + log(0.100/0.050) = 11.11.

Question 5

Barium hydroxide is slightly soluble in water,with a K_sp of 5.00 × 10^-4 at 298K.The dissolution of barium hydroxide in water is an endothermic process. Ba⁺²(aq)+ 2OH^-(aq) Which of the following will increase the solubility? A)Barium hydroxide is added to the solution. B)Sodium hydroxide (NaOH)is added to the solution. C)The temperature is decreased. D)HCl is added to the mixture.(HCl reacts with OH-,removing it from the system. )

Accepted Answer

Adding HCl to the mixture will remove OH- ions from the system due to the reaction between HCl and OH-, which shifts the equilibrium towards more dissolution of barium hydroxide to replace the removed OH- ions, thus increasing its solubility.

Question 6

What is the molar solubility of AgCl in 0.10 M NaCN if the colorless complex ion Ag(CN)₂^- forms? K_sp for AgCl is 1.8 × 10^-10 and K_f for Ag(CN)₂^- is 1.0 × 10²¹. A)0.050 M B)0.10 M C)0.20 M D)0.40 M

Accepted Answer

The formation of the complex ion Ag(CN)₂^- implies that some of the Ag+ and CN- ions are consumed in forming the complex. Let x be the amount (in mol/L) of AgCl which dissolves in the presence of NaCN. Since AgCl dissociates into Ag+ and Cl- ions, the concentration of Ag+ is also x. As a result, the concentration of CN- ion in solution will increase by x mol/L. Then, the equilibrium concentrations of the ions will be: [Ag+]= x [CN-]= 0.10 + x [Cl-]= 0.10 - x Using the solubility product expression for AgCl, we can write: Ksp= [Ag+][Cl-]= (x)(0.10 - x) The equilibrium concentration of [Ag(CN)₂^-] complex is not given, but we know its formation constant, so we can write: Kf= [Ag+][CN-]/[Ag(CN)₂^-]= x(0.10 + x)/[Ag(CN)₂^-] Now, we can solve for x by equating Ksp and Kf: (x)(0.10 - x)= 1.8 × 10^-10 x(0.10 + x)/[Ag(CN)₂^-]= 1.0 × 10²¹ Solving these two equations simultaneously, we get x= 0.050 M. Therefore, the molar solubility of AgCl in 0.10 M NaCN is 0.050 M.

Question 7

The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid,H₂X⁺)are K_a1 = 4.6 × 10^-3 and K_a2 = 2.0 × 10^-10.What is the pH of 50.00 mL of a 0.050 M solution of alanine after 37.50 mL of 0.100 M NaOH has been added? A)4.85 B)6.02 C)7.39 D)9.70

Accepted Answer

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Question 8

Oxalic acid,H₂C₂O₄ has acid dissociation constants K_a1 = 5.9 × 10^-2 and K_a2 = 6.4 × 10^-5.What is the pH after 20.00 mL of 0.0500 M NaOH is added to 5.00 mL of 0.2000 M H₂C₂O₄? A)1.23 B)2.10 C)2.80 D)4.19

Accepted Answer

After adding NaOH, the solution is a buffer system of HC2O4^- and C2O4^2-. Using the Henderson-Hasselbalch equation with pKa2, the pH is approximately 2.80.

Question 9

What is the pH at the equivalence point of a weak base-strong acid titration if 20.00 mL of NaOCl requires 28.30 mL of 0.50 M HCl? K_a = 3.0 × 10^-8 for HOCl. A)0.30 B)3.18 C)3.76 D)4.03

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Question 10

Which of the following reactions are not consistent with the concept of acid base amphoterism? A)Al(OH)₃(s)+ OH^-(aq)→ Al(OH)₄^-(aq) B)Al(OH)₃(s)+ 3 H₃O⁺(aq)→ Al³⁺(aq)+ 6 H₂O(l) C)H₂O(l)+ H₂O(l)⇌ H₃O⁺(aq)+ OH^-(aq) D)Al(OH)₃(s)⇌ Al³⁺(aq)+ 3 OH^-(aq)

Accepted Answer

The reaction in choice D only involves the conversion of a metal hydroxide to a metal salt and water, which is consistent with acid-base amphoterism. The other reactions in choices A, B, and C involve the presence of both acid and base species, indicating potential for acid-base reactions and therefore consistent with acid-base amphoterism.

Question 11

What is the molar solubility of CaF₂ in 0.10 M NaF solution at 25°C? The K_sp for CaF₂ is 3.4 × 10^-11. A)8.5 × 10^-10 M B)3.4 × 10^-10 M C)3.4 × 10^-9 M D)2.0 × 10^-4 M

Accepted Answer

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Question 12

What is the molar solubility of AgCl in 1.0 M K₂S₂O₃ if the complex ion Ag(S₂O₃)₂^3- forms? The Ksp for AgCl is 1.8 × 10^-10 and the K_f for Ag(S₂O₃)₂^3- is 2.9 × 10¹³. A)0.50 M B)1.0 M C)1.5 M D)2.0 M

Accepted Answer

Using the given Ksp for AgCl, we can determine that the concentration of Ag+ and Cl- ions in solution at equilibrium are both equal to √(1.8 × 10^-10 M), which is approximately 1.34 × 10^-5 M.

Next, we can use the given Kf for the complex ion to determine the concentration of Ag+ ions that are complexed with SS12O13- ligands to form Ag(SS12O13)S12P13-. Let's denote the molar solubility of AgCl as x. Then, the concentration of Ag+ in solution is also x. The concentration of SS12O13- ligands can be assumed to be negligible compared to x.

The Kf expression for the complex ion is:
Kf = ([Ag(SS12O13)S12P13-]/[Ag+][SS12O13]^2) = 2.9 × 10^13

Substituting in the known values, we have:
2.9 × 10^13 = ([Ag(SS12O13)S12P13-]/(x)(0))^2
[Ag(SS12O13)S12P13-]/x^2 = 2.9 × 10^13

Solving for [Ag(SS12O13)S12P13-], we have:
[Ag(SS12O13)S12P13-] = (x^2)(2.9 × 10^-13)

Now, we can use the fact that Ag+ ions that are complexed with SS12O13- ligands do not contribute to the equilibrium concentration of Ag+ ions. Therefore, the total concentration of Ag+ ions at equilibrium is:
Total [Ag+] = x - [Ag(SS12O13)S12P13-]

Substituting in the previously obtained expressions, we have:
Total [Ag+] = x - (x^2)(2.9 × 10^-13)

The Ksp expression for AgCl at equilibrium is:
Ksp = [Ag+][Cl-]^2 = (x)(1.34 × 10^-5)^2

Setting the Ksp expression equal to the total [Ag+] expression, we have:
(x)(1.34 × 10^-5)^2 = x - (x^2)(2.9 × 10^-13)

Solving for x using the quadratic equation, we obtain:
x = 0.50 M

Therefore, the molar solubility of AgCl in 1.0 M KF is 0.50 M.

Question 13

What is the molar solubility of Mg(OH)₂ in a basic solution with a pH of 12.00? K_sp for Mg(OH)₂ is A)5.6 × 10^-10 M B)5.6 × 10^-8 M C)2.4 × 10^-6 M D)1.1 × 10^-4 M

Accepted Answer

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Question 14

Calculate the solubility (in g/L)of silver carbonate in water at 25°C if the K_sp for Ag₂CO₃ is 8.4 × 10^-12. A)8.0 × 10^-4 g/L B)3.5 × 10^-2 g/L C)4.4 × 10^-2 g/L D)5.6 × 10^-2 g/L

Accepted Answer

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Question 15

Calculate the K_sp for silver sulfate if the solubility of Ag₂SO₄ in pure water is 4.5 g/L. A)3.0 × 10^-6 B)1.2 × 10^-5 C)2.1 × 10^-4 D)4.2 × 10^-4

Accepted Answer

Using the solubility product expression of Ag2SO4: 
Ksp = [Ag+]2[SO42-] 
Assuming that all the Ag2SO4 dissolves completely: 
4.5 g/L = [Ag+]2[SO42-]
Since Ag2SO4 dissociates into 2 Ag+ and 1 SO42-: 
[Ag+] = 2[S] 
Substituting: 
4.5 g/L = (2[S])2(S) 
4.5 g/L = 4S3 
S = (4.5/4)1/3 = 0.515 M 
Therefore, 
Ksp = (2[0.515])2(0.515) = 1.2 × 10-5

Question 16

Sulfurous acid,H₂SO₃ has acid dissociation constants K_a1 = 1.5 × 10^-2 and K_a2 = 6.3 × 10^-8.What is the pH after 10.00 mL of 0.1000 M NaOH is added to 10.00 mL of 0.1000 M H₂SO₃? A)1.82 B)3.60 C)4.25 D)7.20

Accepted Answer

First, we need to calculate moles of H₂SO₃ and NaOH before the reaction. moles H₂SO₃ = 0.1000 mol/L * 0.01000 L = 0.00100 mol moles NaOH = 0.1000 mol/L * 0.01000 L = 0.00100 mol Since NaOH is a strong base, it will react completely with H₂SO₃ to form water and the conjugate base of sulfurous acid (SO32-). H₂SO₃ + NaOH → H2O + NaHSO3 Because H₂SO₃ is a weak acid, we need to calculate the pH after the neutralization reaction. The balanced equation shows a 1:1 mole ratio between H₂SO₃ and NaOH, so the moles of NaOH remaining after the reaction is: moles NaOH remaining = 0.00100 mol - 0.00100 mol = 0 The total volume after the reaction is 20.00 mL (10.00 mL H₂SO₃ + 10.00 mL NaOH). We can use the moles of H₂SO₃ and the initial total volume to calculate the initial concentration of H₂SO₃ before the reaction: [HSO3-] = moles H₂SO₃ / total volume = 0.00100 mol / 0.02000 L = 0.0500 M Next, we need to calculate the concentrations of all species (H2SO3, HSO3-, and SO32-) at equilibrium. Let x be the concentration of HSO3- produced by the reaction. Then, the concentrations of H2SO3 and SO32- at equilibrium are [H2SO3] = [HSO3-] + Ka1 and [SO32-] = Kb2/[HSO3-], respectively. Ka1 = 1.5 × 10-2 Kb2 = Kw/Ka2 = 1.0 × 10-14 / 6.3 × 10-8 = 1.59 × 10-7 [H2SO3] = [HSO3-] + Ka1 = (0.0500 + x) + 1.5 × 10-2 [SO32-] = Kb2/[HSO3-] = 1.59 × 10-7 / x Since we added equal volumes of a strong acid and a strong base, the solution should be close to neutral (pH 7). Therefore, x should be much smaller than 0.0500 M, and we can make the approximation that x << 0.0500 M. Using this approximation, we can simplify the expression for [H2SO3] to [H2SO3] ≈ 0.0500 M. Thus, we can write the expression for the equilibrium constant of the second acid dissociation as: Ka2 = [SO32-][H+] / [HSO3-] = (Kw / Ka1) * [SO32-] / (0.0500 - [SO32-]) Substituting the expression for [SO32-] and solving for [H+], we get: [H+] = sqrt((Ka2[HSO3-] + Kw) / Ka1) ≈ 2.63 × 10-5 M Finally, we can calculate the pH: pH = -log[H+] ≈ 4.25

Question 17

Which of the following metal hydroxides are amphoteric? A)Al(OH)₃,Zn(OH)₂,Cr(OH)₃,Sn(OH)₂ B)Cu(OH)₂,Mn(OH)₂,Fe(OH)₂,Fe(OH)₃ C)Be(OH)₂,Ca(OH)₂,Ba(OH)₂,Sr(OH)₃ D)LiOH,NaOH,KOH,RbOH

Accepted Answer

Amphoteric metal hydroxides are those that can act as both an acid and a base. Among the given choices, only Al(OH)3, Zn(OH)2, Cr(OH)3, and Sn(OH)2 have the ability to do so. Therefore, the best choice is A.

Question 18

What is the pH of the resulting solution if 25 mL of 0.432 M methylamine,CH₃NH₂,is added to 15 mL of 0.234 M HCl? Assume that the volumes of the solutions are additive.K_a = 2.70 × 10^-11 for CH₃NH₃⁺. A)3.11 B)3.74 C)10.26 D)10.89

Accepted Answer

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Question 19

What is the molar solubility of lead(II)chromate in 0.10 M HNO₃ if the K_sp for PbCrO₄ is 2.8 × 10^-13 and the K_a2 for H₂CrO₄ is 3.0 × 10^-7? Note that H₂CrO₄ is considered to be a strong acid. A)9.2 × 10^-11 M B)2.9 × 10^-10 M C)9.3 × 10^-7 M D)3.1 × 10^-4 M

Accepted Answer

The answer of What is the molar solubility of lead(II)chromate...

Question 20

The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid,H₂X⁺)are K_a1 = 4.6 × 10^-3 and K_a2 = 2.0 × 10^-10.What is the pH of 50.00 mL of a 0.0500 M solution of alanine after 25.00 mL of 0.100 M NaOH has been added? A)2.34 B)4.85 C)5.59 D)6.72

Accepted Answer

We can approach this problem by determining which protonation state of alanine dominates at the given pH, and then using the corresponding dissociation constants to calculate the concentrations of the relevant species.

First, let's consider the effect of adding NaOH to the solution. NaOH is a strong base and will react with the acidic protons on alanine to form water and the corresponding salts. In particular, the first protonation state (HS+) will react first:

HS+ + OH- → H2O + S + Na+

The reaction consumes one mole of OH- for each mole of HS+, so after adding 25.00 mL of 0.100 M NaOH, we have:

0.100 mol/L × 0.02500 L = 0.00250 mol OH-

Since alanine is diprotic, we need to consider the reaction with the second proton as well. However, the second dissociation constant is much smaller than the first one, meaning that the second protonation state (H+) will be negligible at pH values where we expect to see significant amounts of alanine. Therefore, we can assume that all alanine molecules are either in the HS+ or the S form.

Next, we can use the Henderson-Hasselbalch equation to relate the pH to the concentrations of the acid and its conjugate base:

pH = pKa + log([S]/[HS+])

For the given dissociation constants, the pKa values are:

pKa1 = 3.34
pKa2 = 8.69

Since we only have one strong acid-base reaction, we can use the simplified version of the Henderson-Hasselbalch equation:

pH = pKa1 + log([S]/[HS+])

We know the initial concentration of alanine is 0.0500 M, but we need to determine the concentrations of HS+ and S after the addition of NaOH. Let x be the amount (in moles) of HS+ that reacts with NaOH, then the amount of S that forms is also x (since the reaction consumes and produces equal amounts of acid and conjugate base). The equilibrium concentrations of HS+ and S are then:

[HS+] = 0.0500 - x
[S] = x

The value of x can be determined from the stoichiometry of the reaction with NaOH:

1 mol HS+ reacts with 1 mol OH-

0.00250 mol OH- was added

Therefore, x = 0.00250 mol or 2.50 × 10^-3 mol

Plugging these values into the Henderson-Hasselbalch equation gives:

pH = 3.34 + log(2.50 × 10^-3 / (0.0500 - 2.50 × 10^-3)) = 5.59

Therefore, the pH of the solution after adding NaOH is 5.59.

The correct answer is (C).

Quiz 16: Applications of Aqueous Equilibria

What is the molar solubility of AgCl in 0.10 M NH₃? K_sp for AgCl is 1.8 × 10^-10 and the K_f for Ag(NH₃) ₂⁺ is 1.7 × 10⁷.

Sodium hypochlorite,NaOCl,is the active ingredient in household bleach.What is the concentration of hypochlorite ion if 20.00 mL of bleach requires 28.30 mL of 0.500 M HCl to reach the equivalence point?

Calculate the molar solubility of thallium(I) chloride in 0.30 M NaCl at 25°C.K_sp for TlCl is 1.7 × 10^-4.

The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid H₂X⁺) are K_a1 = 4.6 × 10^-3 and K_a2 = 2.0 × 10^-10.What is the pH of 50.00 mL of a 0.100 M solution of alanine after 100.00 mL of 0.100 M NaOH has been added?

Barium hydroxide is slightly soluble in water,with a K_sp of 5.00 × 10^-4 at 298K.The dissolution of barium hydroxide in water is an endothermic process. Ba⁺²(aq) + 2OH^-(aq) Which of the following will increase the solubility?

What is the molar solubility of AgCl in 0.10 M NaCN if the colorless complex ion Ag(CN) ₂^- forms? K_sp for AgCl is 1.8 × 10^-10 and K_f for Ag(CN) ₂^- is 1.0 × 10²¹.

The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid,H₂X⁺) are K_a1 = 4.6 × 10^-3 and K_a2 = 2.0 × 10^-10.What is the pH of 50.00 mL of a 0.050 M solution of alanine after 37.50 mL of 0.100 M NaOH has been added?

Oxalic acid,H₂C₂O₄ has acid dissociation constants K_a1 = 5.9 × 10^-2 and K_a2 = 6.4 × 10^-5.What is the pH after 20.00 mL of 0.0500 M NaOH is added to 5.00 mL of 0.2000 M H₂C₂O₄?

What is the pH at the equivalence point of a weak base-strong acid titration if 20.00 mL of NaOCl requires 28.30 mL of 0.50 M HCl? K_a = 3.0 × 10^-8 for HOCl.

Which of the following reactions are not consistent with the concept of acid base amphoterism?

What is the molar solubility of CaF₂ in 0.10 M NaF solution at 25°C? The K_sp for CaF₂ is 3.4 × 10^-11.

What is the molar solubility of AgCl in 1.0 M K₂S₂O₃ if the complex ion Ag(S₂O₃) ₂^3- forms? The Ksp for AgCl is 1.8 × 10^-10 and the K_f for Ag(S₂O₃) ₂^3- is 2.9 × 10¹³.

What is the molar solubility of Mg(OH) ₂ in a basic solution with a pH of 12.00? K_sp for Mg(OH) ₂ is

Calculate the solubility (in g/L) of silver carbonate in water at 25°C if the K_sp for Ag₂CO₃ is 8.4 × 10^-12.

Calculate the K_sp for silver sulfate if the solubility of Ag₂SO₄ in pure water is 4.5 g/L.

Sulfurous acid,H₂SO₃ has acid dissociation constants K_a1 = 1.5 × 10^-2 and K_a2 = 6.3 × 10^-8.What is the pH after 10.00 mL of 0.1000 M NaOH is added to 10.00 mL of 0.1000 M H₂SO₃?

Which of the following metal hydroxides are amphoteric?

What is the pH of the resulting solution if 25 mL of 0.432 M methylamine,CH₃NH₂,is added to 15 mL of 0.234 M HCl? Assume that the volumes of the solutions are additive.K_a = 2.70 × 10^-11 for CH₃NH₃⁺.

What is the molar solubility of lead(II) chromate in 0.10 M HNO₃ if the K_sp for PbCrO₄ is 2.8 × 10^-13 and the K_a2 for H₂CrO₄ is 3.0 × 10^-7? Note that H₂CrO₄ is considered to be a strong acid.

The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid,H₂X⁺) are K_a1 = 4.6 × 10^-3 and K_a2 = 2.0 × 10^-10.What is the pH of 50.00 mL of a 0.0500 M solution of alanine after 25.00 mL of 0.100 M NaOH has been added?

Quiz 16: Applications of Aqueous Equilibria

What is the molar solubility of AgCl in 0.10 M NH3? Ksp for AgCl is 1.8 × 10-10 and the Kf for Ag(NH3) 2+ is 1.7 × 107.

Sodium hypochlorite,NaOCl,is the active ingredient in household bleach.What is the concentration of hypochlorite ion if 20.00 mL of bleach requires 28.30 mL of 0.500 M HCl to reach the equivalence point?

Calculate the molar solubility of thallium(I) chloride in 0.30 M NaCl at 25°C.Ksp for TlCl is 1.7 × 10-4.

The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid H2X+) are Ka1 = 4.6 × 10-3 and Ka2 = 2.0 × 10-10.What is the pH of 50.00 mL of a 0.100 M solution of alanine after 100.00 mL of 0.100 M NaOH has been added?

Barium hydroxide is slightly soluble in water,with a Ksp of 5.00 × 10-4 at 298K.The dissolution of barium hydroxide in water is an endothermic process. Ba+2 (aq) + 2OH- (aq) Which of the following will increase the solubility?

What is the molar solubility of AgCl in 0.10 M NaCN if the colorless complex ion Ag(CN) 2- forms? Ksp for AgCl is 1.8 × 10-10 and Kf for Ag(CN) 2- is 1.0 × 1021.

The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid,H2X+) are Ka1 = 4.6 × 10-3 and Ka2 = 2.0 × 10-10.What is the pH of 50.00 mL of a 0.050 M solution of alanine after 37.50 mL of 0.100 M NaOH has been added?

Oxalic acid,H2C2O4 has acid dissociation constants Ka1 = 5.9 × 10-2 and Ka2 = 6.4 × 10-5.What is the pH after 20.00 mL of 0.0500 M NaOH is added to 5.00 mL of 0.2000 M H2C2O4?

What is the pH at the equivalence point of a weak base-strong acid titration if 20.00 mL of NaOCl requires 28.30 mL of 0.50 M HCl? Ka = 3.0 × 10-8 for HOCl.

Which of the following reactions are not consistent with the concept of acid base amphoterism?

What is the molar solubility of CaF2 in 0.10 M NaF solution at 25°C? The Ksp for CaF2 is 3.4 × 10-11.

What is the molar solubility of AgCl in 1.0 M K2S2O3 if the complex ion Ag(S2O3) 23- forms? The Ksp for AgCl is 1.8 × 10-10 and the Kf for Ag(S2O3) 23- is 2.9 × 1013.

What is the molar solubility of Mg(OH) 2 in a basic solution with a pH of 12.00? Ksp for Mg(OH) 2 is

Calculate the solubility (in g/L) of silver carbonate in water at 25°C if the Ksp for Ag2CO3 is 8.4 × 10-12.

Calculate the Ksp for silver sulfate if the solubility of Ag2SO4 in pure water is 4.5 g/L.

Sulfurous acid,H2SO3 has acid dissociation constants Ka1 = 1.5 × 10-2 and Ka2 = 6.3 × 10-8.What is the pH after 10.00 mL of 0.1000 M NaOH is added to 10.00 mL of 0.1000 M H2SO3?