Question 1

What is the pH of a solution made by mixing 25.00 mL of 0.100 mol L^-1 HCl with 40.00 mL of 0.100 mol L^-1 KOH? Assume that the volumes of the solutions are additive. A)0.64 B)1.64 C)12.36 D)13.36

Accepted Answer

First, we need to calculate the moles of HCl and KOH:

moles HCl = (0.100 mol/L) x (0.02500 L) = 0.00250 mol
moles KOH = (0.100 mol/L) x (0.04000 L) = 0.00400 mol

Since KOH is a strong base and HCl is a strong acid, they will react completely to form water and a salt (KCl):

HCl + KOH → H2O + KCl

The limiting reagent in this reaction is HCl, which will react completely. The excess KOH will determine the pH of the solution.

The moles of excess KOH are:

moles excess KOH = moles KOH - moles HCl = 0.00400 mol - 0.00250 mol = 0.00150 mol

The concentration of excess KOH is:

[excess KOH] = moles excess KOH / total volume = 0.00150 mol / (0.02500 L + 0.04000 L) = 0.0200 mol/L

Since KOH is a strong base, it will completely dissociate in water to form OH- ions:

KOH → K+ + OH-

The concentration of OH- ions is therefore 0.0200 mol/L.

The pH of the solution can be calculated as follows:

pH = 14 - pOH
pOH = -log[OH-]
pOH = -log(0.0200) = 1.70
pH = 14 - 1.70 = 12.30

Therefore, the pH of the solution is 12.30, which is closest to option C (12.36).

Question 2

Calculate the K_sp for silver sulfite if the solubility of Ag₂SO₃ in pure water is 4.6 × 10^-3g L^-1. A)3.8 × 10^-15 B)1.5 × 10^-14 C)2.4 × 10^-10 D)4.8 × 10^-10

Accepted Answer

The equation for the dissolving of silver sulfite in water is Ag2SO3(S) ↔ 2Ag+(aq) + SO32-(aq).

Using the given solubility of Ag2SO3 in water (4.6 × 10-3 g/L or 4.6 × 10-6 mol/L), we can set up an ICE table:

Initial: 0 mol/L Ag+, 0 mol/L SO32-
Change: +2x mol/L Ag+, +x mol/L SO32-
Equilibrium: 2x mol/L Ag+, x mol/L SO32-

We know that the Ksp expression for Ag2SO3 is [Ag+]2[SO32-]. Plugging in the equilibrium concentrations from the ICE table:

Ksp = (2x)2(x) = 4x3

We need to solve for x, which represents the molar solubility of Ag2SO3.

4.6 × 10-6 mol/L = x
x = 4.6 × 10-6 mol/L

Plugging this value back into the Ksp expression:

Ksp = 4(4.6 × 10-6)3 = 1.5 × 10-14

Therefore, the correct answer is B) 1.5 × 10-14.

Question 3

What is the pH of the resulting solution if 45 mL of 0.432 mol L^-1 methylamine, CH₃NH₂, is added to 15 mL of 0.234 mol L^-1 HCl? Assume that the volumes of the solutions are additive. K_a = 2.70 × 10^-11for CH₃NH₃⁺. A)2.77 B)4.09 C)9.91 D)11.23

Accepted Answer

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

Question 4

How many millilitres of 0.0850 mol L^-1 NaOH are required to titrate 25.0 mL of 0.0720 mol L^-1 HBr to the equivalence point? A)21.2 B)0.245 C)3.92 D)0.153 E)29.5

Accepted Answer

The moles of HBr are 0.0720 mol/L * 0.025 L = 0.0018 mol. Since NaOH and HBr react in a 1:1 ratio, 0.0018 mol of NaOH is needed. The volume of NaOH is 0.0018 mol / 0.0850 mol/L = 0.0212 L, which is 21.2 mL.

Question 5

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 mol L^-1 HCl for complete neutralization? K_a = 3.0 × 10^-8 for HOCl. A)0.30 B)3.18 C)3.76 D)4.03

Accepted Answer

The answer of What is the pH at the equivalence...

Question 6

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH)requires 29.80 mL of 0.3567 mol L^-1 NaOH? K_a =1.8 × 10^-4 for formic acid. A)2.06 B)5.48 C)8.52 D)11.94

Accepted Answer

At the equivalence point, moles of NaOH added = moles of HCOOH initially present. Using the balanced chemical equation for the reaction between NaOH and HCOOH, we can find the number of moles of HCOOH initially present:

HCOOH + NaOH → NaCOOH + H2O

1 mole of HCOOH reacts with 1 mole of NaOH.

0.3567 mol/L x 0.02980 L = 0.01062556 moles of NaOH

Since the volume of HCOOH used is 25 mL = 0.025 L, we can find the initial concentration of HCOOH:

0.01062556 mol / 0.025 L = 0.4250224 mol/L

Now we can use the given acid dissociation constant (Ka) for formic acid to find the pH at the equivalence point. Since HCOOH is a weak acid and NaOH is a strong base, the pH at the equivalence point will be greater than 7 (basic). We can set up the following equilibrium equation:

HCOOH + OH- <--> H2O + HCOO-

At the equivalence point, we have added enough NaOH to completely neutralize all of the HCOOH, so the concentration of HCOOH is now zero. We can use an ICE (Initial, Change, Equilibrium) table to find the concentration of OH- and HCOO-:

Initial: [OH-] = [HCOO-] = 0
Change: [OH-] = x, [HCOO-] = x
Equilibrium: [OH-] = [HCOO-] = x

Since we know the initial concentration of HCOOH and the concentration of HCOO- at equilibrium, we can use the acid dissociation constant (Ka) to find the concentration of OH- at equilibrium:

Ka = [H+][HCOO-] / [HCOOH]

We can assume that the concentration of H+ is negligible compared to the concentration of HCOO-, so:

Ka = [HCOO-]^2 / [HCOOH]

Solving for [HCOO-]:

[HCOO-] = sqrt(Ka[HCOOH])

[HCOO-] = sqrt(1.8 x 10^-4 x 0.4250224)

[HCOO-] = 0.0086948 mol/L

At the equivalence point, the concentration of OH- is equal to the concentration of HCOO-, so:

[OH-] = 0.0086948 mol/L

Taking the negative logarithm (base 10) of this concentration gives the pOH at the equivalence point:

pOH = -log([OH-])

pOH = -log(0.0086948)

pOH = 2.06

Finally, we can find the pH at the equivalence point by subtracting the pOH from 14:

pH = 14 - pOH

pH = 14 - 2.06

pH = 11.94

Therefore, the best choice is C with the explanation provided above.

Question 7

A 25.0 mL sample of 0.150 mol L^-1 hypochlorous acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH after 26.0 mL of base is added? The K_a of hypochlorous acid is 3.0 × 10^-8. A)2.54 B)11.47 C)7.00 D)7.51 E)7.54

Accepted Answer

The answer of A 25.0 mL sample of 0.150 mol...

Question 8

Calculate the solubility (in g L^-1)of calcium fluoride in water at 25 °C if the K_sp for CaF₂ is 1.5 × 10^-10. A)9.6 × 10^-4 g L^-1 B)2.6 × 10^-2 g L^-1 C)3.3 × 10^-2 g L^-1 D)4.1 × 10^-2 g L^-1

Accepted Answer

The solubility product expression for CaF2 is Ksp = [Ca2+][F-]^2. Let the solubility of CaF2 be s mol/L. Then [Ca2+] = s and [F-] = 2s. Ksp = s(2s)^2 = 4s^3. Solving 4s^3 = 1.5 × 10^-10 gives s = 3.42 × 10^-4 mol/L. The molar mass of CaF2 is 78.07 g/mol, so solubility in g/L is 3.42 × 10^-4 mol/L × 78.07 g/mol = 2.67 × 10^-2 g/L, which is approximately 2.6 × 10^-2 g/L.

Question 9

What is the pH of a solution made by mixing 10.00 mL of 0.10 mol L^-1 acetic acid with 10.00 mL of 0.10 mol L^-1 KOH? Assume that the volumes of the solutions are additive. K_a = 1.8 × 10^-5for CH₃COOH. A)5.28 B)7.00 C)8.72 D)10.02

Accepted Answer

First, we need to find the moles of acetic acid and KOH:
moles of CH3COOH = 0.1 mol/L x 0.01 L = 0.001 mol
moles of KOH = 0.1 mol/L x 0.01 L = 0.001 mol

Since KOH is a strong base, it will completely dissociate into its ions:
KOH → K+ + OH-

The OH- ions will react with the CH3COOH to form water and the acetate ion (CH3COO-):
CH3COOH + OH- → CH3COO- + H2O

The initial moles of CH3COOH (0.001 mol) will be reduced by the same amount as the moles of OH- added (0.001 mol). Therefore, the remaining moles of CH3COOH will be:
0.001 mol - 0.001 mol = 0 mol

The moles of CH3COO- formed will be the same as the moles of OH- added (0.001 mol). Therefore, the concentration (in mol/L) of the acetate ion will be:
0.001 mol / 0.02 L = 0.05 mol/L

Now we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log([CH3COO-]/[CH3COOH])

The pKa for acetic acid is 4.76. Substituting in the values, we get:
pH = 4.76 + log(0.05/0.1) = 8.72

Therefore, the pH of the solution is 8.72, which corresponds to answer choice C.

Question 10

What is the pH of a solution made by mixing 30.00 mL of 0.10 mol L^-1 acetic acid (CH₃COOH)with 50.00 mL of 0.100 mol L^-1 KOH? Assume that the volumes of the solutions are additive. K_a = 1.8 × 10^-5for CH₃COOH. A)8.26 B)9.26 C)11.13 D)12.40

Accepted Answer

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

Question 11

What is the pH of the resulting solution if 25.00 mL of 0.10 mol L^-1 acetic acid is added to 10.00 mL of 0.10 mol L^-1 NaOH? Assume that the volumes of the solutions are additive. K_a = 1.8 × 10^-5for CH₃COOH. A)9.43 B)9.08 C)4.92 D)4.57

Accepted Answer

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

Question 12

A 25.0 mL sample of 0.150 mol L^-1 hydrazoic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH after 13.3 mL of base is added? The K_a of hydrazoic acid is 1.9 × 10^-5. A)4.45 B)1.34 C)3.03 D)4.78 E)4.66

Accepted Answer

The answer of A 25.0 mL sample of 0.150 mol...

Question 13

A 25.0 mL sample of 0.150 mol L^-1 hydrofluoric acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH at the equivalence point? The K_a of hydrofluoric acid is 3.5 × 10^-4. A)10.17 B)10.83 C)3.17 D)7.00 E)8.17

Accepted Answer

At the equivalence point, the solution contains the salt NaF, which hydrolyzes to form OH⁻ ions, making the solution basic. The pH is calculated using the Kb of F⁻, which is derived from Kw/Ka. The pH at the equivalence point is approximately 8.17.

Question 14

Calculate the molar solubility of thallium chloride (TlCl)in 0.40 mol L^-1 NaCl at 25 °C. K_sp for TlCl is 1.7 × 10^-4. A)6.8 × 10^-5 mol L^-1 B)4.2 × 10^-4 mol L^-1 C)8.2 × 10^-3 mol L^-1 D)1.3 × 10^-2 mol L^-1

Accepted Answer

We can use the following equation to calculate the molar solubility of TlCl:

Ksp = [Tl+][Cl-]

Let x be the molar solubility of TlCl in NaCl. Then the equilibrium concentration of Tl+ and Cl- will both be equal to x. Therefore:

Ksp = x^2

Substituting the values given in the problem, we get:

1.7 × 10^-4 = x^2

Solving for x, we get:

x = 4.2 × 10^-3 mol L^-1

Therefore, the molar solubility of TlCl in 0.40 mol L^-1 NaCl is:

x/(0.40 mol L^-1) = 4.2 × 10^-3/0.40 = 1.05 × 10^-2 or 4.2 × 10^-4 mol L^-1

The closest choice to this value is B, which is 4.2 × 10^-4 mol L^-1. Therefore, the correct answer is B.

Question 15

What is the molar solubility of Mg(OH)₂ in a basic solution with a pH of 12.50? K_spfor Mg(OH)₂ is 5.6 × 10^-12. A)1.8 × 10^-10 mol L^-1 B)5.6 × 10^-9 mol L^-1 C)2.4 × 10^-6 mol L^-1 D)1.1 × 10^-4 mol L^-1

Accepted Answer

The first step in solving this problem is to write the balanced equation for the dissolution of Mg(OH)2 in water:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

The equilibrium constant expression for this reaction is:
Ksp = [Mg2+][OH-]2

Since Mg(OH)2 is a base, it will partially dissolve in a basic solution (pH > 7). Specifically, the OH- ion concentration will be higher than in pure water, which will lead to an increase in the solubility of Mg(OH)2.

To calculate the molar solubility, we need to find the concentration of Mg2+ and OH- ions in solution. Since the pH is given, we can use the fact that:
pH + pOH = 14

Therefore, pOH = 14 - pH = 1.5.

From this, we can calculate the OH- ion concentration as:
[OH-] = 10^-pOH = 3.2 × 10^-2 M

Using the Ksp expression, we can solve for [Mg2+]:
Ksp = [Mg2+][OH-]2
5.6 × 10^-12 = [Mg2+](3.2 × 10^-2)^2
[Mg2+] = 5.6 × 10^-9 M

Finally, we can calculate the molar solubility as the concentration of Mg(OH)2 in equilibrium with the solution:
Molar solubility = [Mg(OH)2] = [Mg2+] = 5.6 × 10^-9 M

The only answer choice that is closest to this value is B, which is therefore the correct answer.

Question 16

How many millilitres of 0.120 mol L^-1 NaOH are required to titrate 50.0 mL of 0.0998 mol L^-1 butanoic acid to the equivalence point? Butanoic acid is monoprotic. The K_a of butanoic acid is 1.5 × 10^-5. A)4.90 B)50.0 C)41.6 D)60.1 E)4.65

Accepted Answer

The balanced chemical equation for the reaction between NaOH and butanoic acid is:
NaOH + HC4H7O2 → NaC4H7O2 + H2O

From the equation, we can see that 1 mole of NaOH reacts with 1 mole of HC4H7O2. Therefore, the amount of NaOH required to react with 0.0998 mol HC4H7O2 is also 0.0998 mol.

At the equivalence point, the moles of NaOH should be equal to the moles of HC4H7O2.

n(NaOH) = n(HC4H7O2)
0.0998 mol = (C_vol/L_vol) x 0.120 mol
C_vol/L_vol = 0.8317
L_vol = 0.0500 L (given)

Therefore, the volume of NaOH required to react with 50.0 mL of 0.0998 mol/L HC4H7O2 is:

V(NaOH) = (0.8317 mol/L) x 0.0500 L = 0.0416 L or 41.6 mL

The answer is therefore (C) 41.6.

Question 17

Sodium hypochlorite, NaOCl, is the active ingredient in household bleach. What is the concentration of hypochlorite ion if 20.00 mL of bleach requires 32.00 mL of 0.500 mol L^-1 HCl to reach the equivalence point? A)0.300 mol L^-1 B)0.312 mol L^-1 C)0.800 mol L^-1 D)1.30 mol L^-1

Accepted Answer

The answer of Sodium hypochlorite, NaOCl, is the active ingredient...

Question 18

A 25.0 mL sample of 0.150 mol L^-1 butanoic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH before any base is added? Butanoic acid is monoprotic. The K_a of butanoic acid is 1.5 × 10^-5. A)2.83 B)1.5 × 10^-3 C)4.82 D)4.00 E)1.0 × 10⁴

Accepted Answer

The pH before any base is added can be calculated using the formula for the pH of a weak acid: pH = -log(√(Ka × [HA])). Substituting the given values: pH = -log(√(1.5 × 10^-5 × 0.150)) = 2.83.

Question 19

Formic acid (HCOOH, K_a = 1.8 × 10^-4)is the principal component in the venom of stinging ants. What is the molarity of a formic acid solution if 25.00 mL of the formic acid solution requires 29.80 mL of 0.0567 mol L^-1 NaOH to reach the equivalence point? A)0.0134 mol L^-1 B)0.0476 mol L^-1 C)0.0567 mol L^-1 D)0.0676 mol L^-1

Accepted Answer

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Question 20

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous hydrofluoric acid requires 30.00 mL of 0.400 mol L^-1 NaOH? K_a = 6.76 × 10^-4 for HF. A)1.74 B)5.75 C)8.25 D)12.26

Accepted Answer

At the equivalence point of a weak acid-strong base titration, the moles of the strong base added equals the moles of the weak acid present. Therefore, we can use the balanced chemical equation of the reaction to determine the number of moles of HF present in the initial solution.

From the balanced equation: 
1 mol HF reacts with 1 mol NaOH 
0.400 mol NaOH reacts with (0.400 mol / 1 L) x (0.03000 L) = 0.012 mol HF

Therefore, 25 mL of aqueous hydrofluoric acid contains 0.012 mol HF: 
(0.012 mol / 0.025 L) = 0.480 mol/L HF

Using the expression for the ionization constant of HF: 
pH = pKa + log([A-]/[HA]) 
pH = pKa + log([NaF]/[HF])

At the equivalence point, [NaF] = [HF] because the salt NaF is formed by the complete neutralization of HF. Therefore,

pH = pKa + log(1) 
pH = pKa

Substituting in the pKa value of HF: 
pH = -log(Ka) = -log(6.76 x 10^-4) 
pH = 8.25

Therefore, the approximate pH at the equivalence point of this titration is 8.25, which corresponds to answer choice C.

Quiz 16: Aqueous Ionic Equilibrium

What is the pH of a solution made by mixing 25.00 mL of 0.100 mol L^-1 HCl with 40.00 mL of 0.100 mol L^-1 KOH? Assume that the volumes of the solutions are additive.

Calculate the K_sp for silver sulfite if the solubility of Ag₂SO₃ in pure water is 4.6 × 10^-3g L^-1.

What is the pH of the resulting solution if 45 mL of 0.432 mol L^-1 methylamine, CH₃NH₂, is added to 15 mL of 0.234 mol L^-1 HCl? Assume that the volumes of the solutions are additive. K_a = 2.70 × 10^-11for CH₃NH₃⁺.

How many millilitres of 0.0850 mol L^-1 NaOH are required to titrate 25.0 mL of 0.0720 mol L^-1 HBr to the equivalence point?

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 mol L^-1 HCl for complete neutralization? K_a = 3.0 × 10^-8 for HOCl.

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH) requires 29.80 mL of 0.3567 mol L^-1 NaOH? K_a =1.8 × 10^-4 for formic acid.

A 25.0 mL sample of 0.150 mol L^-1 hypochlorous acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH after 26.0 mL of base is added? The K_a of hypochlorous acid is 3.0 × 10^-8.

Calculate the solubility (in g L^-1) of calcium fluoride in water at 25 °C if the K_sp for CaF₂ is 1.5 × 10^-10.

What is the pH of a solution made by mixing 10.00 mL of 0.10 mol L^-1 acetic acid with 10.00 mL of 0.10 mol L^-1 KOH? Assume that the volumes of the solutions are additive. K_a = 1.8 × 10^-5for CH₃COOH.

What is the pH of a solution made by mixing 30.00 mL of 0.10 mol L^-1 acetic acid (CH₃COOH) with 50.00 mL of 0.100 mol L^-1 KOH? Assume that the volumes of the solutions are additive. K_a = 1.8 × 10^-5for CH₃COOH.

What is the pH of the resulting solution if 25.00 mL of 0.10 mol L^-1 acetic acid is added to 10.00 mL of 0.10 mol L^-1 NaOH? Assume that the volumes of the solutions are additive. K_a = 1.8 × 10^-5for CH₃COOH.

A 25.0 mL sample of 0.150 mol L^-1 hydrazoic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH after 13.3 mL of base is added? The K_a of hydrazoic acid is 1.9 × 10^-5.

A 25.0 mL sample of 0.150 mol L^-1 hydrofluoric acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH at the equivalence point? The K_a of hydrofluoric acid is 3.5 × 10^-4.

Calculate the molar solubility of thallium chloride (TlCl) in 0.40 mol L^-1 NaCl at 25 °C. K_sp for TlCl is 1.7 × 10^-4.

What is the molar solubility of Mg(OH) ₂ in a basic solution with a pH of 12.50? K_spfor Mg(OH) ₂ is 5.6 × 10^-12.

How many millilitres of 0.120 mol L^-1 NaOH are required to titrate 50.0 mL of 0.0998 mol L^-1 butanoic acid to the equivalence point? Butanoic acid is monoprotic. The K_a of butanoic acid is 1.5 × 10^-5.

Sodium hypochlorite, NaOCl, is the active ingredient in household bleach. What is the concentration of hypochlorite ion if 20.00 mL of bleach requires 32.00 mL of 0.500 mol L^-1 HCl to reach the equivalence point?

A 25.0 mL sample of 0.150 mol L^-1 butanoic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH before any base is added? Butanoic acid is monoprotic. The K_a of butanoic acid is 1.5 × 10^-5.

Formic acid (HCOOH, K_a = 1.8 × 10^-4) is the principal component in the venom of stinging ants. What is the molarity of a formic acid solution if 25.00 mL of the formic acid solution requires 29.80 mL of 0.0567 mol L^-1 NaOH to reach the equivalence point?

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous hydrofluoric acid requires 30.00 mL of 0.400 mol L^-1 NaOH? K_a = 6.76 × 10^-4 for HF.

Quiz 16: Aqueous Ionic Equilibrium

What is the pH of a solution made by mixing 25.00 mL of 0.100 mol L-1 HCl with 40.00 mL of 0.100 mol L-1 KOH? Assume that the volumes of the solutions are additive.

Calculate the Ksp for silver sulfite if the solubility of Ag2SO3 in pure water is 4.6 × 10-3 g L-1.

What is the pH of the resulting solution if 45 mL of 0.432 mol L-1 methylamine, CH3NH2, is added to 15 mL of 0.234 mol L-1 HCl? Assume that the volumes of the solutions are additive. Ka = 2.70 × 10-11 for CH3NH3+.

How many millilitres of 0.0850 mol L-1 NaOH are required to titrate 25.0 mL of 0.0720 mol L-1 HBr to the equivalence point?

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 mol L-1 HCl for complete neutralization? Ka = 3.0 × 10-8 for HOCl.

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH) requires 29.80 mL of 0.3567 mol L-1 NaOH? Ka =1.8 × 10-4 for formic acid.

A 25.0 mL sample of 0.150 mol L-1 hypochlorous acid is titrated with a 0.150 mol L-1 NaOH solution. What is the pH after 26.0 mL of base is added? The Ka of hypochlorous acid is 3.0 × 10-8.

Calculate the solubility (in g L-1) of calcium fluoride in water at 25 °C if the Ksp for CaF2 is 1.5 × 10-10.

What is the pH of a solution made by mixing 10.00 mL of 0.10 mol L-1 acetic acid with 10.00 mL of 0.10 mol L-1 KOH? Assume that the volumes of the solutions are additive. Ka = 1.8 × 10-5 for CH3COOH.

What is the pH of a solution made by mixing 30.00 mL of 0.10 mol L-1 acetic acid (CH3COOH) with 50.00 mL of 0.100 mol L-1 KOH? Assume that the volumes of the solutions are additive. Ka = 1.8 × 10-5 for CH3COOH.

What is the pH of the resulting solution if 25.00 mL of 0.10 mol L-1 acetic acid is added to 10.00 mL of 0.10 mol L-1 NaOH? Assume that the volumes of the solutions are additive. Ka = 1.8 × 10-5 for CH3COOH.

A 25.0 mL sample of 0.150 mol L-1 hydrazoic acid is titrated with a 0.150 mol L-1 NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of hydrazoic acid is 1.9 × 10-5.

A 25.0 mL sample of 0.150 mol L-1 hydrofluoric acid is titrated with a 0.150 mol L-1 NaOH solution. What is the pH at the equivalence point? The Ka of hydrofluoric acid is 3.5 × 10-4.

Calculate the molar solubility of thallium chloride (TlCl) in 0.40 mol L-1 NaCl at 25 °C. Ksp for TlCl is 1.7 × 10-4.

What is the molar solubility of Mg(OH) 2 in a basic solution with a pH of 12.50? Ksp for Mg(OH) 2 is 5.6 × 10-12.

How many millilitres of 0.120 mol L-1 NaOH are required to titrate 50.0 mL of 0.0998 mol L-1 butanoic acid to the equivalence point? Butanoic acid is monoprotic. The Ka of butanoic acid is 1.5 × 10-5.

Sodium hypochlorite, NaOCl, is the active ingredient in household bleach. What is the concentration of hypochlorite ion if 20.00 mL of bleach requires 32.00 mL of 0.500 mol L-1 HCl to reach the equivalence point?

A 25.0 mL sample of 0.150 mol L-1 butanoic acid is titrated with a 0.150 mol L-1 NaOH solution. What is the pH before any base is added? Butanoic acid is monoprotic. The Ka of butanoic acid is 1.5 × 10-5.

Formic acid (HCOOH, Ka = 1.8 × 10-4) is the principal component in the venom of stinging ants. What is the molarity of a formic acid solution if 25.00 mL of the formic acid solution requires 29.80 mL of 0.0567 mol L-1 NaOH to reach the equivalence point?

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous hydrofluoric acid requires 30.00 mL of 0.400 mol L-1 NaOH? Ka = 6.76 × 10-4 for HF.