Question 1

What volume of 5.00 × 10^-3 M HNO₃ is needed to titrate 100.00 mL of 5.00 × 10^-3 M Ca(OH)₂ to the equivalence point? A)12.5 mL B)50.0 mL C)100.mL D)200.mL

Accepted Answer

The answer of What volume of 5.00 × 10^-3 M...

Question 2

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

Accepted Answer

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

Question 3

What is the pH at the equivalence point of a weak acid-strong base titration?
A)pH < 7
B)pH = 7
C)pH > 7
D)pH = 14.00

Accepted Answer

At the equivalence point of a weak acid-strong base titration, the strong base completely neutralizes the weak acid to produce the conjugate base of the weak acid and water. Since the conjugate base of a weak acid is a weak base, it will slightly hydrolyze in water to produce hydroxide ions (OH-) which will increase the pH of the solution. Therefore, the pH at the equivalence point of a weak acid-strong base titration will be greater than 7, making choice C the correct answer.

Question 4

At what pH is the amino acid glycine with a K_a of 2.51 × 10^-10 sixty-six (66%)percent dissociated? A)9)60 B)9)89 C)10.10 D)10.60

Accepted Answer

For an amino acid like glycine to be 66% dissociated, it means that the pH is at a point where the concentration of the dissociated form is twice that of the undissociated form. This is because at 66% dissociation, 2/3 of the molecules are dissociated, leaving 1/3 undissociated, giving a ratio of 2:1. This situation corresponds to the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the dissociated form and [HA] is the concentration of the undissociated form. When [A-] is twice [HA], log([A-]/[HA]) = log(2) = 0.3. Given that the pKa (which is the -log of the given dissociation constant, KS) is 9.60 (since pKa = -log(2.51 × 10^-10)), the pH at which glycine is 66% dissociated is 9.60 + 0.3 = 9.90, which rounds to 9.89.

Question 5

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.The addition of NaOH to alanine will result in the deprotonation of the protonated form of alanine, H2X+, to form the anionic form of alanine, X-. The pH of the solution will increase as a result of the decrease in the concentration of H+ ions.The equilibrium constants for the dissociation of H2X+ are:Ka1 = 4.6 × 10^-3Ka2 = 2.0 × 10^-10The pH of the solution can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log([A-]/[HA])where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the anionic form of the acid, and [HA] is the concentration of the protonated form of the acid.In this case, the pKa of the protonated form of alanine is:pKa1 = -log(4.6 × 10^-3) = 2.34The concentration of the anionic form of alanine can be calculated using the following equation:[A-] = [NaOH] - [H2X+]where [NaOH] is the concentration of NaOH added and [H2X+] is the concentration of the protonated form of alanine.In this case, [NaOH] = 0.100 M and [H2X+] can be calculated using the following equation:[H2X+] = Ka1[A-]/[OH-]where [OH-] is the concentration of hydroxide ions.In this case, [OH-] can be calculated using the following equation:[OH-] = Kw/[H+]where Kw is the ion product of water (1.0 × 10^-14).In this case, [H+] can be calculated using the following equation:[H+] = 10^-pHSubstituting the appropriate values into the above equations, we get:[A-] = 0.100 M - 4.6 × 10^-3 M = 0.0954 M[H2X+] = 4.6 × 10^-3 M * 0.0954 M / 1.0 × 10^-14 M = 4.36 × 10^-12 MpH = 2.34 + log(0.0954 M / 4.36 × 10^-12 M) = 11.11Therefore, the pH of the solution is 11.11.

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 requires 29.80 mL of 0.0567 M NaOH? K_a = 1.8 × 10^-4for formic acid. A)2)46 B)5)88 C)8)12 D)11.54

Accepted Answer

At the equivalence point of a weak acid-strong base titration, the conjugate base of the weak acid is present in solution. In this case, formic acid (HCOOH) is the weak acid and its conjugate base is formate (HCOO-).

The balanced equation for the reaction is: HCOOH + NaOH -> HCOONa + H2O

Using stoichiometry, the moles of NaOH used can be calculated:
(0.0567 mol/L) x (0.02980 L) = 0.00168826 mol NaOH

Since the acid and the base are in a 1:1 ratio, the moles of HCOOH used is also 0.00168826 mol.

Using the equilibrium expression for the acid dissociation constant (Ka), the concentration of H+ can be calculated: 
Ka = [H+][HCOO-]/[HCOOH]
1.8 x 10^-4 = x^2/(0.025 L - 0.02980 L)
x = [H+] = 0.0044 M

Since pH = -log[H+], the pH at the equivalence point is approximately 8.12. Therefore, the best choice is C.

Question 7

Which of the following titrations result in a basic solution at the equivalence point? A)HI titrated with NaCH₃CO₂ B)HOCl titrated with NaOH C)HBr titrated with KOH D)Pb(NO₃)₂ titrated with NaI

Accepted Answer

HOCl is a weak acid and NaOH is a strong base, so the equivalence point will result in a basic solution.

Question 8

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

The balanced equation for the reaction between NaOH and H2SO3 is:

NaOH + H2SO3 → NaHSO3 + H2O

First, we need to calculate the moles of H2SO3 present in the solution:

moles H2SO3 = (10.00 mL)(0.1000 M) = 0.001000 mol H2SO3

Since NaOH reacts with H2SO3 in a 1:1 ratio, the number of moles of NaOH added is also 0.001000 mol.

The first step in determining the pH of the solution is to calculate the initial concentrations of HS2O3- and H2SO3. We can assume that all of the dissolved H2SO3 is initially in the form of HS2O3-, so the concentration of HS2O3- (initially) is:

[HS2O3-] = moles HS2O3- / total volume
[HS2O3-] = (0.001000 mol) / 0.02000 L
[HS2O3-] = 0.05000 M

The concentration of H2SO3 (initially) can be calculated using the acid dissociation constant Ka1:

Ka1 = [H+][HSO3-] / [H2SO3]
1.5 × 10^-2 = x^2 / (0.05000 - x)
Assuming that x << 0.05000, we can simplify this expression to:

1.5 × 10^-2 = x^2 / 0.05000

Solving for x, we get:

x = 3.87 × 10^-4 M

Therefore, the concentration of H2SO3 (initially) is:

[H2SO3] = 3.87 × 10^-4 M

Next, we need to determine how much NaOH was consumed in the reaction:

0.001000 mol NaOH - 0.001000 mol H2SO3 = 0.000000 mol NaOH

Since no NaOH was consumed, the remaining concentration of NaOH is still 0.1000 M.

The remaining amount of HS2O3- and H2SO3 can be determined using an ICE (initial-change-equilibrium) table:

|             | HS2O3-  | H2SO3   |
|-------------|--------|---------|
| Initial     | 0.0500 | 0.000387|
| Change      | -x     | +x      |
| Equilibrium | 0.0500-x| 0.000387+x|

Since the reaction consumes H2SO3, the value of x must be less than 0.000387:

0.0500 - x = [HSO3-]
0.0500 - x = sqrt(Ka2*([HS2O3-]))
[HS2O3-] / [HSO3-] = 6.3 × 10^-14

Solving these two equations simultaneously yields:

[HSO3-] = 7.29 × 10^-7 M
[HS2O3-] = 0.0493 M

The total concentration of sulfurous acid is given by:

[HSO3-] + [HS2O3-] = 0.0493 + 7.29 × 10^-7
[HSO3-] + [HS2O3-] = 0.0493 M

To calculate the pH, we can use the following expression:

pH = pKa + log([A-] / [HA])

We will use pKa2 (since we calculated the concentrations of both HS2O3- and HSO3-) and assume that [A-] = [HSO3-] and [HA] = [HS2O3-]:

pH = 1.17 + log(7.29 × 10^-7 / 0.0493)
pH = 4.25

Therefore, the pH after 10.00 mL of 0.1000 M NaOH is added to 10.00 mL of 0.1000 M HS2O3- is 4.25. The answer is C.

Question 9

The equivalence point pH of the titration of four weak acids is given.Which is the strongest acid?
A)7)24
B)7)86
C)8)12
D)8)43

Accepted Answer

The answer of The equivalence point pH of the titration...

Question 10

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

The answer of The dissociation equilibrium constants for the protonated...

Question 11

Which of the following titrations result in an acidic solution at the equivalence point? A)CH₃COOH titrated with NaOH B)KF titrated with KOH C)HCl titrated with NaOH D)C₅H₅N titrated with HCl

Accepted Answer

The titration of pyridine (C5H5N), a weak base, with HCl, a strong acid, results in an acidic solution at the equivalence point.

Question 12

What is the pH of the resulting solution if 30.00 mL of 0.10 M acetic acid is added to 10.00 mL of 0.10 M NaOH? Assume that the volumes of the solutions are additive.K_a = 1.8 × 10^-5for CH₃CO₂H. A)9)56 B)8)95 C)5)05 D)4)44

Accepted Answer

First, we need to write out the balanced chemical equation for the reaction between acetic acid and sodium hydroxide: 
CH3COOH + NaOH → CH3COONa + H2O 
From this equation, we can see that the acid and base will neutralize each other, with the resulting solution containing the sodium acetate salt and water. 
To determine the pH of the resulting solution, we need to calculate the moles of acid and base present, and then use the amount of sodium acetate formed (which is the same as the amount of acid that reacted) to calculate the concentration of acetate ions in the solution. From there, we can use the equilibrium constant (Ka) for acetic acid to calculate the pH. 
Here are the steps:

1. Calculate the moles of acid and base:

moles of acid = 0.030 L * 0.10 mol/L = 0.0030 mol 
moles of base = 0.010 L * 0.10 mol/L = 0.0010 mol

2. Determine the limiting reactant:

Since the acid and base have a 1:1 stoichiometry, whichever is present in lower moles will be the limiting reactant. In this case, the base is limiting, since there are fewer moles of it present.

3. Calculate the amount of sodium acetate formed:

Since the base is limiting, all of its moles will react with an equal number of moles of acid. Therefore, the amount of sodium acetate formed is equal to the amount of acid that reacted:

moles of acetate = 0.0030 mol

4. Calculate the concentration of acetate ions:

We know that 0.0030 mol of sodium acetate is present in a total solution volume of 40.00 mL (30.00 mL + 10.00 mL), which is 0.0400 L:

[acetate] = moles of acetate / total volume 
[acetate] = 0.0030 mol / 0.0400 L 
[acetate] = 0.075 M

5. Use the Ka for acetic acid to calculate the pH:

Ka = [H+][acetate] / [acid] 
We can assume that the [H+] comes entirely from the dissociation of water (since the acetate ion is a weak base and so will not significantly hydrolyze):

Ka = [H+] * [acetate] / [acid] 
Solving for [H+]:

[H+] = Ka * [acid] / [acetate] 
[H+] = 1.8 × 10^-5 * 0.0030 mol / 0.075 mol 
[H+] = 7.2 × 10^-7 M

Finally, we convert the H+ ion concentration to pH:

pH = -log[H+] 
pH = -log(7.2 × 10^-7) 
pH = 6.14

However, since we are dealing with dilute solutions, we need to take into account the effect of the volume of the NaOH solution on the final pH. We can do this by using the Henderson-Hasselbalch equation, which takes into account the initial concentrations of the acid and base, as well their ratio (which determines the pKa value used in the equation). The Henderson-Hasselbalch equation is:

pH = pKa + log([base] / [acid])

The pKa value for acetic acid is 4.74. Plugging in our values:

pH = 4.74 + log(0.010 / 0.040) 
pH = 4.74 - 0.70 
pH = 4.04

Therefore, our final answer is:

Answer: D 
 The pH of the resulting solution is approximately 4.04.

Question 13

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

First, let's write out the dissociation reactions for alanine:
HS1U1B12S1U1B0XS1U1P1+S1S1P0 ⇆ H+ + HS1U1B11S1U1B0XS1U1P1+S1S1P0
HS1U1B11S1U1B0XS1U1P1+S1S1P0 ⇆ H+ + S1U1B10XS1U1P1+S1S1P0
We can set up an ICE table for each reaction.
For the first reaction:
HS1U1B12S1U1B0XS1U1P1+S1S1P0 ⇆ H+ + HS1U1B11S1U1B0XS1U1P1+S1S1P0
I:    0.050 M         0              0  
C: -x              +x           +x
E: 0.050-x        x              x
Ks1 = 4.6 × 10S1U1P1-3S1S1P0
Ks1 = [H+][HS1U1B11S1U1B0XS1U1P1+S1S1P0]/[HS1U1B12S1U1B0XS1U1P1+S1S1P0]
4.6 × 10S1U1P1-3S1S1P0 = xS1U1P1S1S1P0 / (0.050 - x)S1U1P1S1S1P0
Assumption: x is negligible compared to 0.050, so 0.050 - x ≈ 0.050
4.6 × 10S1U1P1-3S1S1P0 = xS1U1P1S1S1P0 / 0.050
x = 1.15 × 10S1U1P1-S1S1P0 M = [H+]
Now we can calculate the pH:
pH = -log[H+] = -log(1.15 × 10S1U1P1-S1S1P0) = 5.59
Next, we need to consider the second dissociation. However, since the equilibrium constant for KS1U1B1a2S1U1B0 is much smaller than that for KS1U1B1aS1U1B0, we can assume that the concentration of the fully deprotonated form is negligible compared to the concentration of the singly deprotonated form. Therefore, we can ignore the second dissociation in this problem.
Next, we need to consider the addition of NaOH. NaOH reacts with H+ to form water, which means that we are effectively removing H+ ions from the solution. Since we initially had a solution with excess H+ ions, the addition of NaOH will cause the pH to rise, but not above the pKa1 of alanine.
The balanced equation for the reaction between NaOH and H+ is:
NaOH + H+ → Na+ + H2O
We can use stoichiometry to determine how many moles of H+ ions are neutralized by 25.00 mL of 0.100 M NaOH:
0.100 M NaOH × 0.02500 L = 2.5 × 10S1U1P1-S1S1P0 mol NaOH
Since NaOH and H+ react in a 1:1 ratio, we neutralize 2.5 × 10S1U1P1-S1S1P0 mol H+ ions.
What is the concentration of H+ ions remaining after the addition of NaOH?
We started with x = 1.15 × 10S1U1P1-S1S1P0 M H+ ions, but we just neutralized 2.5 × 10S1U1P1-S1S1P0 mol H+ ions. Therefore, the number of moles of H+ ions remaining is:
x - 2.5 × 10S1U1P1-S1S1P0 mol H+ ions
The total volume of the solution is now 50.00 mL + 25.00 mL = 75.00 mL = 0.07500 L.
Therefore, the new concentration of H+ ions is:
(x - 2.5 × 10S1U1P1-S1S1P0 mol) / 0.07500 L = (x - 3.33 × 10S1U1P1-S1S1P0) M
pH = -log[H+] = -log(x - 3.33 × 10S1U1P1-S1S1P0)
pH = -log(1.15 × 10S1U1P1-S1S1P0 - 3.33 × 10S1U1P1-S1S1P0) = 5.59. 
Therefore, the final answer is choice C, 5.59.

Question 14

What is the percent dissociation of ascorbic acid if the solution has a pH = 5.50 and a pK_a = 4.10? A)96% B)10% C)5% D)1%

Accepted Answer

The percent dissociation of ascorbic acid can be calculated using the formula for percent dissociation given the pH and pKa of the solution. The formula is: % dissociation = 100 / (1 + 10^(pKa - pH)). Plugging in the given values (pKa = 4.10, pH = 5.50), we get % dissociation = 100 / (1 + 10^(4.10 - 5.50)) = 96%. This indicates that 96% of the ascorbic acid is dissociated at a pH of 5.50.

Question 15

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

The balanced equation for the reaction between hypochlorite ion and HCl is:

Hypochlorite ion + HCl --> Chloride ion + H2O + Cl2

From the equation, we can see that 1 mole of HCl reacts with 1 mole of hypochlorite ion. Therefore, the number of moles of hypochlorite ion in 28.30 mL of 0.500 M HCl is:

moles of HCl = M * V = 0.500 M * 0.02830 L = 0.0142 moles
moles of hypochlorite ion = moles of HCl = 0.0142 moles

Since the volume of the bleach used is not given, we assume that the bleach and HCl react in a 1:1 stoichiometric ratio. Therefore, the number of moles of hypochlorite ion in the bleach is also 0.0142 moles.

The concentration of hypochlorite ion can be calculated as follows:

concentration of hypochlorite ion = moles of hypochlorite ion / volume of bleach used
= 0.0142 moles / 0.02000 L
= 0.71 M

However, in bleach, the hypochlorite ion exists in equilibrium with hypochlorous acid according to the following reaction:

Hypochlorite ion + H+ --> Hypochlorous acid

Since the pH of household bleach is usually around 12, most of the hypochlorite ion exists as the ion (OCl-) rather than the acid (HOCl). The equilibrium constant for the reaction is:

K = [OCl-] / [HOCl]

Assuming that most of the hypochlorite ion exists as the ion, we can approximate the concentration of hypochlorite ion as equal to the concentration of the bleach (since the concentration of the acid will be negligible). Therefore, the concentration of hypochlorite ion is approximately 0.71 M, which corresponds to choice C.

Question 16

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

The answer of Oxalic acid,H₂C₂O₄ has acid dissociation constants K_a1...

Question 17

The equivalence point pH of the titration of four weak bases is given.Which is the strongest base?
A)4)21
B)5)67
C)6)33
D)7)58

Accepted Answer

The strongest base will have the highest equivalence point pH, indicating it has the greatest ability to accept protons.

Question 18

Formic acid (HCO₂H,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 M NaOH to reach the equivalence point? A)0)0134 M B)0)0476 M C)0)0567 M D)0)0676 M

Accepted Answer

The answer of Formic acid (HCO₂H,K_a = 1.8 × 10^-4)is...

Question 19

What is the pH of a solution made by mixing 30.00 mL of 0.10 M HCl with 40.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive.
A)0)85
B)1)85
C)12.15
D)13.15

Accepted Answer

This is a neutralization reaction between HCl and KOH, which produces water and a salt (KCl). The balanced equation is:

HCl + KOH → H2O + KCl

To find the pH, we need to determine the concentration of the resulting solution. First, we can calculate the number of moles of each reactant:

moles of HCl = 0.10 M x 0.03000 L = 0.00300 mol
moles of KOH = 0.10 M x 0.04000 L = 0.00400 mol

Since KOH is the limiting reactant (it will be completely consumed), we can use its moles to calculate the concentration of the resulting solution:

moles of KCl = 0.00400 mol
volume of resulting solution = 0.03000 L + 0.04000 L = 0.07000 L

[ KCl ] = moles / volume = 0.00400 mol / 0.07000 L = 0.057 M

Now we can use the dissociation of water to find the pH:

Kw = [H3O+][OH-] = 1.0 x 10^-14

In a neutral solution, [H3O+] = [OH-], so:

[H3O+] = [OH-] = sqrt(Kw) = 1.0 x 10^-7 M

Since the solution contains KCl, it will have no effect on pH (it is a neutral salt). Therefore:

pH = -log[H3O+] = -log(1.0 x 10^-7) = 7.00

However, we must also consider the effect of the HCl and KOH on the pH. HCl is a strong acid that completely dissociates in water, so it contributes H3O+ ions to the solution. KOH is a strong base that also completely dissociates in water, so it contributes OH- ions to the solution. The net effect will depend on the amounts of HCl and KOH:

moles of HCl > moles of KOH: The excess H3O+ ions will react with the OH- ions from KOH to form water. The resulting solution will be acidic, with a pH less than 7.

moles of HCl < moles of KOH: The excess OH- ions will react with the H3O+ ions from HCl to form water. The resulting solution will be basic, with a pH greater than 7.

moles of HCl = moles of KOH: The H3O+ and OH- ions will completely neutralize each other, forming water. The resulting solution will be neutral, with a pH of 7.

In this case, we have:

moles of HCl = 0.00300 mol
moles of KOH = 0.00400 mol

Therefore, KOH is the limiting reactant, and there are excess OH- ions. The H3O+ and OH- ions will combine to form water. The resulting solution will be basic, with a pH greater than 7. To calculate the pH, we need to find the concentration of excess OH- ions:

moles of OH- in excess = moles of KOH - moles of HCl = 0.00400 mol - 0.00300 mol = 0.00100 mol

[OH-] = moles / volume = 0.00100 mol / 0.07000 L = 0.014 M

pOH = -log[OH-] = -log(0.014) = 1.85

pH = 14.00 - pOH = 14.00 - 1.85 = 12.15

Therefore, the pH of the solution is 12.15. The correct answer is (C).

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 M 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 base added will equal the moles of acid present in the solution. Using the given information, we can calculate the moles of NaOH used by multiplying the concentration (0.400 M) by the volume used (30.00 mL), then dividing by 1000 to convert to liters:

moles NaOH = (0.400 M) x (30.00 mL/1000) = 0.0120 mol

Since NaOH is a strong base, we know that it will completely dissociate into Na+ and OH- ions in water. Therefore, we can calculate the concentration of OH- ions present in the solution at the equivalence point by dividing the moles of NaOH by the total volume of the solution (which is the sum of the volumes of the acid and base used):

total volume = 25 mL + 30.00 mL = 55.00 mL = 0.0550 L

[OH-] = (0.0120 mol) / (0.0550 L) = 0.218 M

Now we can use the equilibrium constant expression for the dissociation of HF to calculate the concentration of H+ ions present in the solution at the equivalence point:

KS1U1B1aS1U1B0 = [H+][F-] / [HF]

[HF] = moles HF / total volume
moles HF = moles NaOH = 0.0120 mol
[F-] = [Na+] = 0.400 M (due to complete dissociation of NaOH)
[H+] = KS1U1B1aS1U1B0 x [HF] / [F-]
[H+] = (6.76 x 10S1U1P1-4S1S1P0) x (0.0120 mol/0.0550 L) / 0.400 M
[H+] = 1.47 x 10S1U1P0-4S1S1P0 M

Finally, we can use the pH equation to calculate the pH of the solution:

pH = -log[H+]
pH = -log(1.47 x 10S1U1P0-4S1S1P0) = 3.83

Therefore, the approximate pH at the equivalence point of this titration is 3.83, which is closest to choice C (8.25).

Quiz 15: Applications of Aqueous Equilibria

What volume of 5.00 × 10^-3 M HNO₃ is needed to titrate 100.00 mL of 5.00 × 10^-3 M Ca(OH) ₂ 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 M HCl? K_a = 3.0 × 10^-8 for HOCl.

What is the pH at the equivalence point of a weak acid-strong base titration?

At what pH is the amino acid glycine with a K_a of 2.51 × 10^-10 sixty-six (66%) percent dissociated?

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?

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid requires 29.80 mL of 0.0567 M NaOH? K_a = 1.8 × 10^-4for formic acid.

Which of the following titrations result in a basic solution at the equivalence point?

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?

The equivalence point pH of the titration of four weak acids is given.Which is the strongest 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 0.050 M solution of alanine after 37.50 mL of 0.100 M NaOH has been added?

Which of the following titrations result in an acidic solution at the equivalence point?

What is the pH of the resulting solution if 30.00 mL of 0.10 M acetic acid is added to 10.00 mL of 0.10 M NaOH? Assume that the volumes of the solutions are additive.K_a = 1.8 × 10^-5for CH₃CO₂H.

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?

What is the percent dissociation of ascorbic acid if the solution has a pH = 5.50 and a pK_a = 4.10?

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?

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?

The equivalence point pH of the titration of four weak bases is given.Which is the strongest base?

Formic acid (HCO₂H,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 M NaOH to reach the equivalence point?

What is the pH of a solution made by mixing 30.00 mL of 0.10 M HCl with 40.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive.

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 M NaOH? K_a = 6.76 × 10^-4 for HF.

Quiz 15: Applications of Aqueous Equilibria

What volume of 5.00 × 10-3 M HNO3 is needed to titrate 100.00 mL of 5.00 × 10-3 M Ca(OH) 2 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 M HCl? Ka = 3.0 × 10-8 for HOCl.

What is the pH at the equivalence point of a weak acid-strong base titration?

At what pH is the amino acid glycine with a Ka of 2.51 × 10-10 sixty-six (66%) percent dissociated?

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?

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid requires 29.80 mL of 0.0567 M NaOH? Ka = 1.8 × 10-4 for formic acid.