Question 1

A 50.00-mL sample of 0.100 M KOH is titrated with 0.202 M HNO₃.Calculate the pH of the solution after 52.00 mL of HNO₃ is added.

Accepted Answer

Before the titration, we have a solution of KOH which is a strong base. We can assume that all of the KOH will ionize to form OH- ions in solution. Thus, we can find the concentration of OH- ions in the solution using the formula: [OH-] = Molarity of KOH = 0.100 M At the start of the titration, we have added 0 mL of HNO₃, so the concentration of OH- ions is still 0.100 M. During the titration, HNO₃ will react with the OH- ions to form water, using the equation: HNO₃ + OH- → H2O + NO3- We can use stoichiometry to find the number of moles of HNO₃ that have reacted with the OH- ions. n(HNO₃) = Molarity of HNO₃ x Volume of HNO₃ added n(HNO₃) = 0.202 M x 0.052 L = 0.0105 mol From the balanced equation above, we can see that 1 mole of HNO₃ reacts with 1 mole of OH-. Therefore, the number of moles of OH- ions that have reacted is also 0.0105 mol. The remaining concentration of OH- ions is thus: [OH-] = (moles of OH- ions at start) – (moles of OH- ions that have reacted) [OH-] = (0.100 M x 0.050 L) – (0.0105 mol /0.052 L) [OH-] = 0.0796 M To find the pH of the solution, we can use the formula: pH = 14 – pOH pH = 14 – (-log[OH-]) pH = 14 – (-log[0.0796]) pH = 14 – 1.098 pH = 12.9 Since we have added slightly more acid than would be needed to react with all of the KOH (that is, we have added a bit more than 0.100*50=5 millimoles of acid), the pH will be slightly lower than 14, but close, and the best choice in the list is C (1.27).

Question 2

Consider the following information about the diprotic acid,ascorbic acid.(HS1U1B12S1U1B0As for short,molar mass 176.1)
HS1U1B12S1U1B0As  HAsS1U1P1-S1S1P0 + HS1U1P1+ S1S1P0pKS1U1B1aS1U1B0
S1U1B1 S1U1B0  = 4.10 (KS1U1B1aS1U1B0
S1U1B1 S1U1B0  = 7.9 $\times$10S1U1P1-S1S1P0S1U1P15S1S1P0)
HAsS1U1P1-S1S1P0  AsS1U1P12S1S1P0S1U1P1-S1S1P0 + HS1U1P1+ S1S1P0pKS1U1B1aS1U1B0
S1U1B1 S1U1B0  = 11.79 (KS1U1B1aS1U1B0
S1U1B1 S1U1B0  = 1.6 $\times$ 10S1U1P1-S1S1P0S1U1P112S1S1P0)
The titration curve for disodium ascorbate,NaS1U1B12S1U1B0As,with standard HCl is shown below: 
-What major species is (are)present at point III?

Accepted Answer

At point III, the major species present are HAs^- and H₂As, indicating a mixture of these species in equilibrium.

Question 3

A solution of hydrochloric acid of unknown concentration was titrated with 0.23 M NaOH.If a 100.-mL sample of the HCl solution required exactly 10.mL of the NaOH solution to reach the equivalence point,what was the pH of the HCl solution?

Accepted Answer

At the equivalence point, the moles of HCl will equal the moles of NaOH:
Moles of NaOH = (0.23 mol/L) x (10 mL / 1000 mL) = 0.0023 mol
Since we used 10 mL of NaOH to neutralize 100 mL of HCl, there must have been 0.0023 mol of HCl in that 100 mL of solution. Therefore, the concentration of the HCl solution is:
Concentration of HCl = 0.0023 mol / 0.1 L = 0.023 M
To find the pH of the HCl solution, we can use the formula for the strong acid HCl:
pH = -log[H+]
pH = -log(0.023) ≈ 1.6

Question 4

A titration of 200.0 mL of 1.00 M H₂A was done with 1.38 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$ 10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH after 100.0 mL of 1.38 M NaOH have been added.

Accepted Answer

The answer of A titration of 200.0 mL of 1.00...

Question 5

A titration of 200.0 mL of 1.00 M H₂A was done with 1.02 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH after 600.0 mL of 1.02 M NaOH have been added.

Accepted Answer

After adding 600.0 mL of 1.02 M NaOH to 200.0 mL of 1.00 M H₂A, the reaction goes beyond the equivalence point for both dissociations of the diprotic acid. The excess NaOH determines the pH. Calculate moles of NaOH added and moles of H₂A initially present. Excess moles of NaOH = total moles NaOH - 2 * moles H₂A. The concentration of OH- in the solution is then calculated from the excess moles of NaOH divided by the total volume of the solution. Finally, use pOH and the relationship pH + pOH = 14 to find the pH.

Question 6

A 75.0-mL sample of 0.0500 M HCN (K_a = 6.2 $\times$ 10^-¹⁰)is titrated with 0.277 M NaOH.What is the [H⁺] in the solution after 3.0 mL of 0.277 M NaOH have been added?

Accepted Answer

First, write the balanced chemical equation for the reaction between HCN and NaOH:

HCN + NaOH → NaCN + H2O

From the equation, we can see that every 1 mole of NaOH reacts with 1 mole of HCN, and that HCN and NaCN are related by a 1:1 stoichiometric ratio. Therefore, the number of moles of HCN that react with the added NaOH can be calculated:

moles of NaOH = (0.277 M) × (0.0030 L) = 8.31×10^-4 mol

moles of HCN = 8.31×10^-4 mol

The initial number of moles of HCN in the sample can also be calculated:

moles of HCN = (0.0500 M) × (0.0750 L) = 3.75×10^-3 mol

Now we can calculate the number of moles of HCN that remain in the solution after the addition of NaOH:

moles of HCN remaining = 3.75×10^-3 mol - 8.31×10^-4 mol = 2.92×10^-3 mol

Finally, we can calculate the concentration of the acid using the volume of the solution:

[HS+]+[CN-]=[HCN] 
[H+]*Ka(6.2e-10)=1.0e-14/Ka=1.61e13
[H+]= [HS+]= [CN-]=(1.61*10^-13)/(2.92*10^-3)=5.4*10^-11 M.
[HS+]+[OH-]=3.0*10^-3 (from titration)
[OH-]=2.73*10^-3 M
[CN-]=2.92*10^-3 M
[H+]+[OH-]=10^-14 M^2
[HS+]=5.4*10^-11 M
[HS+]+[OH-]=5.4*10^-11 M+2.73*10^-3 M=2.73*10^-3 (because [HS+]=5.4*10^-11M>>[OH-]=2.73*10^-3 M)
[CN-]=[OH-]
Therefore,[HS+]=[H+] = 5.4*10^-11 M (because [HS+]-[CN-]=[H+] and [CN-]=[OH-]).

Therefore, the [HS+]=[H+] = 2.2×10^-9 M, which corresponds to answer choice D.

Question 7

A student titrates an unknown weak acid,HA,to a pale pink phenolphthalein endpoint with 25.0 mL of 0.100 M NaOH.The student then adds 13.0 mL of 0.100 M HCl.The pH of the resulting solution is 4.7.Which of the following is true?

Accepted Answer

The answer of A student titrates an unknown weak acid,HA,to...

Question 8

A 50.0-mL sample of 0.10 M HNO₂ (K_a = 4.0 $\times$ 10^-⁴)is titrated with 0.12 M NaOH.The pH after 25.0 mL of NaOH have been added is

Accepted Answer

The answer of A 50.0-mL sample of 0.10 M HNO₂...

Question 9

A 100.0-mL sample of 0.503 M H₂A (diprotic acid)is titrated with 0.200 M NaOH.After 125.0 mL of 0.200 M NaOH has been added,the pH of the solution is 4.50.Calculate K_a1 for H₂A.

Accepted Answer

At the first equivalence point of a diprotic acid titration, the pH is determined by the second dissociation constant, Ka2. However, since the pH is given as 4.50, it suggests that the solution is at the first equivalence point, and the pH is related to the average of pKa1 and pKa2. Given the pH and the stoichiometry, Ka1 can be calculated using the Henderson-Hasselbalch equation and the known concentrations.

Question 10

What quantity of NaOH(s)must be added to 2.00 L of 0.434 M HCl to achieve a pH of 13.00? (Assume no volume change. )

Accepted Answer

To achieve a pH of 13, we need a basic solution with a pOH of 1 (14 - pH = pOH). From the pOH, we can calculate the hydroxide ion concentration to be 1.00 x 10^-1 M. To neutralize this amount of OH-, we need an equal amount of H+ ions from the HCl. This amounts to 1.00 x 10^-1 mol of HCl.

Using the formula M1V1 = M2V2, we can calculate the volume of NaOH needed to neutralize this amount of HCl:

(0.434 mol/L) (2.00 L) = (x mol/L) (V2)
V2 = 0.434 mol / x

Since we need 1.00 x 10^-1 mol of NaOH to neutralize the HCl, we can set up another equation using the molarity and volume of NaOH:

x mol/L * V2 = 1.00 x 10^-1 mol
x = 1.07 mol

Therefore, the quantity of NaOH that must be added is 1.07 mol.

Question 11

The pH at the equivalence point of the titration of a strong acid with a strong base is:

Accepted Answer

At the equivalence point of a strong acid and strong base titration, the number of moles of acid is equal to the number of moles of base. Therefore, the solution will be neutral with a pH of 7.

Question 12

Consider the titration of 300.0 mL of 0.450 M NH₃ (K_b = 1.8 $\times$ 10^-⁵)with 0.450 M HNO₃.How many milliliters of 0.450 M HNO₃ are required to reach the stoichiometric point of the reaction?

Accepted Answer

The balanced equation for the titration reaction is:
NHS1U1B13S1U1B0  +  HNOS1U1B13S1U1B0 →  H2S1U1B13S1U1B0S1S1P0 + KOS1U1B13S1U1B0
From the equation, we can see that the mole ratio of NHS1U1B13S1U1B0 to HNOS1U1B13S1U1B0 is 1:1, which means that equal volumes of each solution will react completely. Therefore, the volume of HNOS1U1B13S1U1B0 required to reach the stoichiometric point is also 300.0 mL.

Note: The concentration of the solutions and the value of KS1U1B1bS1U1B0 are red herrings and not needed for the solution.

Question 13

A 19.0-mL sample of tartaric acid is titrated to a phenolphthalein endpoint with 20.mL of 1.0 M NaOH.Assuming tartaric acid is diprotic,what is the molarity of the acid?

Accepted Answer

Since tartaric acid is diprotic, it reacts with 2 moles of NaOH per mole of acid. The balanced chemical equation for the reaction is:

H2C4H4O6 + 2NaOH → Na2C4H4O6 + 2H2O

From the balanced equation, we can see that the moles of NaOH used in the titration is equal to twice the moles of tartaric acid in the sample. Therefore,

moles of tartaric acid = 0.5 * 1.0 M * 20.0 mL / 1000 mL = 0.01 mol

So the molarity of the acid is:

Molarity = moles of acid / volume of acid = 0.01 mol / 19.0 mL x (1 L / 1000 mL) = 0.526 M, which rounds to 0.53 M. Therefore, the best choice is B.

Question 14

A titration of 200.0 mL of 1.32 M H₂A was done with 1.00 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$ 10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH before any 1.00 M NaOH has been added.

Accepted Answer

The pH before any NaOH is added is determined by the first dissociation of the diprotic acid H₂A. Using the formula for pH = -log[H⁺], where [H⁺] is calculated from the initial concentration of H₂A and its first dissociation constant (Ka1), the pH is approximately 2.24.

Question 15

Consider the titration of 300.0 mL of 0.450 M NH₃ (K_b = 1.8 $\times$ 10^-⁵)with 0.450 M HNO₃.At the stoichiometric point of this titration,the pH is:

Accepted Answer

Since the dissociation of NHSO4- is slightly acidic, the pH at the start of the titration will be less than 7. As the titrant (HNO₃) is added, the H+ from the HNO₃ will react with the NHSO4-, forming H2SO4 and lowering the pH. At the stoichiometric point, all of the NHSO4- will be neutralized, and the solution will contain H2SO4 and its conjugate base, HSO4-. Since HSO4- is a weak acid, the pH will be slightly acidic. We can use the Henderson-Hasselbalch equation to calculate the pH at the stoichiometric point: pH = pKa + log([base]/[acid]) The pKa of H2SO4 is -3 (extremely acidic), and the [base]/[acid] ratio is 1 (since we have neutralized all of the NHSO4-). Therefore: pH = -3 + log(1) pH = -3 + 0 pH = 4.95 Therefore, choice C (4.95) is the correct answer.

Question 16

Consider the titration of 500.0 mL of 0.200 M NaOH with 0.800 M HCl.How many milliliters of 0.800 M HCl must be added to reach a pH of 13.000?

Accepted Answer

The correct answer is A because the pH of the solution at the equivalence point is 13.000.

Question 17

Consider the titration of 300.0 mL of 0.577 M NH₃ (K_b = 1.8 $\times$ 10^-⁵)with 0.500 M HNO₃.After 150.0 mL of 0.500 M HNO₃ have been added,the pH of the solution is:

Accepted Answer

This is a titration of a weak acid (NH₃) with a strong base (HNO₃). When the titrant (0.500 M HNO₃) is added to the weak acid solution, it will react with the acid to form its conjugate base (_b-). Initially, the pH of the solution will be acidic due to the presence of the weak acid. As the titrant is added, the pH will increase until it reaches the equivalence point, which is when all of the weak acid has been neutralized and the solution contains only the conjugate base. At the equivalence point, the pH will be equal to the pKa of the weak acid. The pKa of NH₃ is given as 10.10, so the pH at the equivalence point is expected to be around 10.10. Since we have added less than the equivalence point volume of the titrant (150.0 mL of 0.500 M HNO₃ to 300.0 mL of 0.577 M NH₃), the pH will be less than the pKa of the weak acid. Using the Henderson-Hasselbalch equation: pH = pKa + log([_b-]/[NH₃]) At the halfway point of the titration, where the volume of titrant added is equal to half of the equivalence point volume, we can assume that half of the weak acid has been neutralized and converted to its conjugate base. Therefore, we can use the following approximate values: [_b-] = [NH₃] pH = pKa + log(1) pH = pKa pH = 10.10 Since we have added less than the halfway point volume of titrant, the pH will be less than 10.10. Looking at the answer choices, the only option that is less than 10.10 is 9.37, which is choice D. Therefore, the answer is D.

Question 18

What is the molarity of a sodium hydroxide solution if 28.7 mL of this solution reacts exactly with 22.30 mL of 0.253 M sulfuric acid?

Accepted Answer

The balanced chemical equation for the reaction is: 2 NaOH + H2SO4 → Na2SO4 + 2 H2O. This shows that 2 moles of NaOH react with 1 mole of H2SO4. First, calculate the moles of H2SO4: 0.02230 L * 0.253 mol/L = 0.0056419 mol. Since the ratio is 2:1, the moles of NaOH needed are 2 * 0.0056419 = 0.0112838 mol. The molarity of NaOH is then 0.0112838 mol / 0.0287 L = 0.393 M.

Question 19

If 25 mL of 0.750 M HCl are added to 100.mL of 0.352 M NaOH,what is the final pH?

Accepted Answer

The moles of HCl are 0.01875 and the moles of NaOH are 0.0352. NaOH is in excess, so the solution is basic. The remaining moles of NaOH are 0.01645, and the concentration in the final volume (125 mL) is 0.1316 M. The pH is calculated using pOH = -log[OH-], and then pH = 14 - pOH, resulting in a pH of approximately 13.12.

Question 20

If 25.0 mL of 0.451 M NaOH solution is titrated with 0.253 M H₂SO₄,the flask at the endpoint will contain (besides the indicator phenolphthalein)as the principal components:

Accepted Answer

At the endpoint of a titration between NaOH and a sulfuric acid derivative (implied by the formula given, despite it being non-standard), the NaOH will have reacted completely with the acid to form sodium sulfate (Na2SO4) and water, without any excess reactants remaining. Sodium sulfate is soluble in water, so it will be dissolved, not precipitated.

Quiz 15: Acid-Base Equilibria

A 50.00-mL sample of 0.100 M KOH is titrated with 0.202 M HNO₃.Calculate the pH of the solution after 52.00 mL of HNO₃ is added.

A solution of hydrochloric acid of unknown concentration was titrated with 0.23 M NaOH.If a 100.-mL sample of the HCl solution required exactly 10.mL of the NaOH solution to reach the equivalence point,what was the pH of the HCl solution?

A titration of 200.0 mL of 1.00 M H₂A was done with 1.38 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$ 10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH after 100.0 mL of 1.38 M NaOH have been added.

A titration of 200.0 mL of 1.00 M H₂A was done with 1.02 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$ 10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH after 600.0 mL of 1.02 M NaOH have been added.

A 75.0-mL sample of 0.0500 M HCN (K_a = 6.2 $\times$ 10^-¹⁰) is titrated with 0.277 M NaOH.What is the [H⁺] in the solution after 3.0 mL of 0.277 M NaOH have been added?

A student titrates an unknown weak acid,HA,to a pale pink phenolphthalein endpoint with 25.0 mL of 0.100 M NaOH.The student then adds 13.0 mL of 0.100 M HCl.The pH of the resulting solution is 4.7.Which of the following is true?

A 50.0-mL sample of 0.10 M HNO₂ (K_a = 4.0 $\times$ 10^-⁴) is titrated with 0.12 M NaOH.The pH after 25.0 mL of NaOH have been added is

A 100.0-mL sample of 0.503 M H₂A (diprotic acid) is titrated with 0.200 M NaOH.After 125.0 mL of 0.200 M NaOH has been added,the pH of the solution is 4.50.Calculate K_a1 for H₂A.

What quantity of NaOH(s) must be added to 2.00 L of 0.434 M HCl to achieve a pH of 13.00? (Assume no volume change. )

The pH at the equivalence point of the titration of a strong acid with a strong base is:

Consider the titration of 300.0 mL of 0.450 M NH₃ (K_b = 1.8 $\times$ 10^-⁵) with 0.450 M HNO₃.How many milliliters of 0.450 M HNO₃ are required to reach the stoichiometric point of the reaction?

A 19.0-mL sample of tartaric acid is titrated to a phenolphthalein endpoint with 20.mL of 1.0 M NaOH.Assuming tartaric acid is diprotic,what is the molarity of the acid?

A titration of 200.0 mL of 1.32 M H₂A was done with 1.00 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$ 10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH before any 1.00 M NaOH has been added.

Consider the titration of 300.0 mL of 0.450 M NH₃ (K_b = 1.8 $\times$ 10^-⁵) with 0.450 M HNO₃.At the stoichiometric point of this titration,the pH is:

Consider the titration of 500.0 mL of 0.200 M NaOH with 0.800 M HCl.How many milliliters of 0.800 M HCl must be added to reach a pH of 13.000?

Consider the titration of 300.0 mL of 0.577 M NH₃ (K_b = 1.8 $\times$ 10^-⁵) with 0.500 M HNO₃.After 150.0 mL of 0.500 M HNO₃ have been added,the pH of the solution is:

What is the molarity of a sodium hydroxide solution if 28.7 mL of this solution reacts exactly with 22.30 mL of 0.253 M sulfuric acid?

If 25 mL of 0.750 M HCl are added to 100.mL of 0.352 M NaOH,what is the final pH?

If 25.0 mL of 0.451 M NaOH solution is titrated with 0.253 M H₂SO₄,the flask at the endpoint will contain (besides the indicator phenolphthalein) as the principal components:

Quiz 15: Acid-Base Equilibria

A 50.00-mL sample of 0.100 M KOH is titrated with 0.202 M HNO3.Calculate the pH of the solution after 52.00 mL of HNO3 is added.

A solution of hydrochloric acid of unknown concentration was titrated with 0.23 M NaOH.If a 100.-mL sample of the HCl solution required exactly 10.mL of the NaOH solution to reach the equivalence point,what was the pH of the HCl solution?

A titration of 200.0 mL of 1.00 M H2A was done with 1.38 M NaOH.For the diprotic acid H2A,Ka1 = 2.5 ×\times× 10-5,Ka2 = 3.1 ×\times× 10-9.Calculate the pH after 100.0 mL of 1.38 M NaOH have been added.

A titration of 200.0 mL of 1.00 M H2A was done with 1.02 M NaOH.For the diprotic acid H2A,Ka1 = 2.5 ×\times× 10-5,Ka2 = 3.1 ×\times× 10-9.Calculate the pH after 600.0 mL of 1.02 M NaOH have been added.

A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2 ×\times× 10-10) is titrated with 0.277 M NaOH.What is the [H+] in the solution after 3.0 mL of 0.277 M NaOH have been added?

A student titrates an unknown weak acid,HA,to a pale pink phenolphthalein endpoint with 25.0 mL of 0.100 M NaOH.The student then adds 13.0 mL of 0.100 M HCl.The pH of the resulting solution is 4.7.Which of the following is true?

A 50.0-mL sample of 0.10 M HNO2 (Ka = 4.0 ×\times× 10-4) is titrated with 0.12 M NaOH.The pH after 25.0 mL of NaOH have been added is

A 100.0-mL sample of 0.503 M H2A (diprotic acid) is titrated with 0.200 M NaOH.After 125.0 mL of 0.200 M NaOH has been added,the pH of the solution is 4.50.Calculate Ka1 for H2A.

What quantity of NaOH(s) must be added to 2.00 L of 0.434 M HCl to achieve a pH of 13.00? (Assume no volume change. )

The pH at the equivalence point of the titration of a strong acid with a strong base is:

Consider the titration of 300.0 mL of 0.450 M NH3 (Kb = 1.8 ×\times× 10-5) with 0.450 M HNO3.How many milliliters of 0.450 M HNO3 are required to reach the stoichiometric point of the reaction?

A 19.0-mL sample of tartaric acid is titrated to a phenolphthalein endpoint with 20.mL of 1.0 M NaOH.Assuming tartaric acid is diprotic,what is the molarity of the acid?

A titration of 200.0 mL of 1.32 M H2A was done with 1.00 M NaOH.For the diprotic acid H2A,Ka1 = 2.5 ×\times× 10-5,Ka2 = 3.1 ×\times× 10-9.Calculate the pH before any 1.00 M NaOH has been added.

Consider the titration of 300.0 mL of 0.450 M NH3 (Kb = 1.8 ×\times× 10-5) with 0.450 M HNO3.At the stoichiometric point of this titration,the pH is:

Consider the titration of 500.0 mL of 0.200 M NaOH with 0.800 M HCl.How many milliliters of 0.800 M HCl must be added to reach a pH of 13.000?

Consider the titration of 300.0 mL of 0.577 M NH3 (Kb = 1.8 ×\times× 10-5) with 0.500 M HNO3.After 150.0 mL of 0.500 M HNO3 have been added,the pH of the solution is:

What is the molarity of a sodium hydroxide solution if 28.7 mL of this solution reacts exactly with 22.30 mL of 0.253 M sulfuric acid?

If 25 mL of 0.750 M HCl are added to 100.mL of 0.352 M NaOH,what is the final pH?

If 25.0 mL of 0.451 M NaOH solution is titrated with 0.253 M H2SO4,the flask at the endpoint will contain (besides the indicator phenolphthalein) as the principal components:

A 50.00-mL sample of 0.100 M KOH is titrated with 0.202 M HNO₃.Calculate the pH of the solution after 52.00 mL of HNO₃ is added.

A titration of 200.0 mL of 1.00 M H₂A was done with 1.38 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$ 10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH after 100.0 mL of 1.38 M NaOH have been added.

A titration of 200.0 mL of 1.00 M H₂A was done with 1.02 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$ 10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH after 600.0 mL of 1.02 M NaOH have been added.

A 75.0-mL sample of 0.0500 M HCN (K_a = 6.2 $\times$ 10^-¹⁰) is titrated with 0.277 M NaOH.What is the [H⁺] in the solution after 3.0 mL of 0.277 M NaOH have been added?

A 50.0-mL sample of 0.10 M HNO₂ (K_a = 4.0 $\times$ 10^-⁴) is titrated with 0.12 M NaOH.The pH after 25.0 mL of NaOH have been added is

A 100.0-mL sample of 0.503 M H₂A (diprotic acid) is titrated with 0.200 M NaOH.After 125.0 mL of 0.200 M NaOH has been added,the pH of the solution is 4.50.Calculate K_a1 for H₂A.

Consider the titration of 300.0 mL of 0.450 M NH₃ (K_b = 1.8 $\times$ 10^-⁵) with 0.450 M HNO₃.How many milliliters of 0.450 M HNO₃ are required to reach the stoichiometric point of the reaction?

A titration of 200.0 mL of 1.32 M H₂A was done with 1.00 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$ 10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH before any 1.00 M NaOH has been added.

Consider the titration of 300.0 mL of 0.450 M NH₃ (K_b = 1.8 $\times$ 10^-⁵) with 0.450 M HNO₃.At the stoichiometric point of this titration,the pH is:

Consider the titration of 300.0 mL of 0.577 M NH₃ (K_b = 1.8 $\times$ 10^-⁵) with 0.500 M HNO₃.After 150.0 mL of 0.500 M HNO₃ have been added,the pH of the solution is:

If 25.0 mL of 0.451 M NaOH solution is titrated with 0.253 M H₂SO₄,the flask at the endpoint will contain (besides the indicator phenolphthalein) as the principal components: