Question 1

A solution containing 10. mmol of CO₃²^- and 5.0 mmol of HCO₃^- is titrated with 1.0 M HCl. What total volume of HCl must be added to reach the second equivalence point?

Accepted Answer

The answer of A solution containing 10. mmol of CO₃²^-...

Question 2

A 200.0-mL sample of the weak acid HS1U1B13S1U1B0A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.)
-After 220.0 mL of 0.200 M NaOH is added

Accepted Answer

The answer of A 200.0-mL sample of the weak acid...

Question 3

A 0.210-g sample of an acid (molar mass = 192 g/mol) is titrated with 30.5 mL of 0.108 M NaOH to a phenolphthalein endpoint. The formula of the acid is

Accepted Answer

The acid reacts with NaOH in a 1:1 molar ratio, so moles of NaOH = moles of acid. Moles of NaOH = 0.108 M * 0.0305 L = 0.003294 moles. The molar mass of the acid is given as 192 g/mol, so moles of acid = 0.210 g / 192 g/mol = 0.001094 moles. The discrepancy in moles suggests the acid is polyprotic and fully reacts with 3 moles of NaOH per mole of acid, indicating the formula is H3A, which corresponds to choice A.

Question 4

A solution contains 10. mmol of H₃PO₄ and 5.0 mmol of NaH₂PO₄. How many milliliters of 0.10 M NaOH must be added to reach the second equivalence point of the titration of the H₃PO₄ with NaOH?

Accepted Answer

To reach the second equivalence point, all H₃PO₄ must be converted to Na₂HPO₄. This requires 2 moles of NaOH per mole of H₃PO₄. Thus, 20 mmol of NaOH is needed. With 0.10 M NaOH, 250 mL is required (20 mmol / 0.10 M = 0.20 L = 250 mL).

Question 5

Calculate the pH when 200.0 mL of a 1.00 M solution of HS1U1B12S1U1B0A (KS1U1B1a1S1U1B0 = 1.0 * 10S1U1P1-S1S1P0S1U1P16S1S1P0, KS1U1B1a2S1U1B0 = 1.0 * 10S1U1P1-S1S1P0S1U1P110S1S1P0) is titrated with the following volumes of 1.00 M NaOH.
-150.0 mL of 1.00 M NaOH

Accepted Answer

Step 1: Write the balanced equation for the reaction between HS1U1B12S1U1B0A and NaOH.
HS1U1B12S1U1B0A + NaOH → NaS1U1P2S1U1P0 + H2O

Step 2: Calculate the number of moles of HS1U1B12S1U1B0A initially present.
moles HS1U1B12S1U1B0A = volume of HS1U1B12S1U1B0A × concentration of HS1U1B12S1U1B0A
moles HS1U1B12S1U1B0A = 0.200 L × 1.00 mol/L = 0.200 mol

Step 3: Since the stoichiometry of the reaction is 1:1 between HS1U1B12S1U1B0A and NaOH, the number of moles of NaOH added to reach the equivalence point is also 0.200 mol.

Step 4: Calculate the volume of NaOH required to reach the equivalence point.
moles NaOH = concentration of NaOH × volume of NaOH
0.200 mol = 1.00 mol/L × volume of NaOH
volume of NaOH = 0.200 L or 200 mL

Step 5: Calculate the remaining volume of NaOH after adding 150.0 mL of 1.00 M NaOH.
remaining volume of NaOH = total volume of NaOH - volume of NaOH used
remaining volume of NaOH = 200 mL - 150.0 mL = 50.0 mL

Step 6: Convert the remaining volume of NaOH to moles.
moles NaOH remaining = concentration of NaOH × volume of NaOH
moles NaOH remaining = 1.00 mol/L × 0.0500 L = 0.0500 mol

Step 7: Calculate the total volume of the solution.
total volume = volume of HS1U1B12S1U1B0A + volume of NaOH used
total volume = 0.200 L + 0.150 L = 0.350 L

Step 8: Calculate the concentration of the NaS1U1P2S1U1P0 ion.
moles NaS1U1P2S1U1P0 = moles HS1U1B12S1U1B0A at equivalence point = 0.200 mol
concentration NaS1U1P2S1U1P0 = moles NaS1U1P2S1U1P0 / total volume
concentration NaS1U1P2S1U1P0 = 0.200 mol / 0.350 L = 0.571 M

Step 9: Calculate the pOH at the equivalence point.
pOH = -log[OH-]
[OH-] = Kw / [H+]
At the equivalence point, [HS1U1B12S1U1B0A] = [NaS1U1P2S1U1P0], therefore assume [HS1U1B12S1U1B0A] = [H+]
Ka1 = 1.00 × 10^-10
[H+] = sqrt(Ka1 × [HS1U1B12S1U1B0A])
[H+] = sqrt(1.00 × 10^-10 × 0.200) = 1.41 × 10^-6 M
[OH-] = Kw / [H+] = 1.00 × 10^-14 / 1.41 × 10^-6 = 7.09 × 10^-9 M
pOH = -log(7.09 × 10^-9) = 8.15

Step 10: Calculate the pH at the equivalence point.
pH = 14.00 - pOH = 14.00 - 8.15 = 5.85

Step 11: Calculate the pH after adding 50.0 mL of 1.00 M NaOH.
moles of NaOH reacting with HS1U1B12S1U1B0A = 0.150 mol
moles of NaOH reacting with NaS1U1P2S1U1P0 = 0.0500 mol
total moles of NaOH = 0.150 mol + 0.0500 mol = 0.200 mol
remaining moles of NaOH = 0.200 mol - 0.200 mol = 0 mol
Therefore, all the HS1U1B12S1U1B0A is consumed and the solution is a 0.571 M solution of NaS1U1P2S1U1P0.
pOH = -log([OH-]) = -log(1.00 × 10^-14/0.571) = 8.48
pH = 14.00 - 8.48 = 5.52

The pH of the solution after adding 150.0 mL of NaOH is at the equivalence point and is calculated to be 5.85. The pH of the solution after adding 50.0 mL of NaOH is above the equivalence point and is calculated to be 5.52. The only answer choice that is within this range is B) 6.48. Therefore, the best choice is B) 6.48.

Question 6

A 5.95-g sample of an acid, H₂X, requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (removing both protons). What is the molar mass of the acid expressed in grams per mol?

Accepted Answer

The answer of A 5.95-g sample of an acid, H₂X,...

Question 7

A 135.0-mL sample of a 0.25 M solution of H₃PO₄ is titrated with 0.12 M NaOH. What volume of base must be added to reach the third equivalence point?

Accepted Answer

The titration of a triprotic acid requires three equivalence points. The third equivalence point represents the point at which all three protons have been neutralized by the added base. In order to calculate the volume of base required to reach this point, we need to first calculate the moles of acid present in the initial solution. moles of acid = Molarity x Volume (in liters) moles of acid = 0.25 M x 0.135 L moles of acid = 0.03375 mol H₃PO₄ Since there are three protons in H₃PO₄, the number of moles of base required to reach the third equivalence point will be three times the number of moles of acid. moles of base required = 3 x moles of acid moles of base required = 3 x 0.03375 mol moles of base required = 0.10125 mol NaOH Now we can use the concentration of NaOH and the number of moles required to calculate the volume of base needed to reach the third equivalence point. volume of base required = moles of base required / Molarity of NaOH volume of base required = 0.10125 mol / 0.12 M volume of base required = 0.84375 L = 843.75 mL The answer is 8.4 F1F1F1S1F1F1F10 10² mL.

Question 8

A 200.0-mL sample of the weak acid HS1U1B13S1U1B0A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.)
-After 75.0 mL of 0.200 M NaOH is added

Accepted Answer

At the beginning of the titration, the major species is the weak acid H₃A. After adding 75.0 mL of 0.200 M NaOH, we have added 0.200 M x 0.0750 L = 0.015 mol of NaOH. The balanced chemical equation for the reaction is: H₃A + NaOH → Na + H2O + H₂A At this point, the amount of NaOH added is less than the amount needed to reach the equivalence point, so the reaction is not complete. However, the addition of NaOH will begin to deprotonate the weak acid. Therefore, the major species at this point will be a mixture of the weak acid H₃A and its conjugate base H₂A^-. The major species does not include the fully deprotonated conjugate base HA²^-, which would only be formed after the equivalence point is reached.

Question 9

Consider the following information about the diprotic acid ascorbic acid (HS1U1B12S1U1B0As for short, molar mass = 176.1). The titration curve for disodium ascorbate, NaS1U1B12S1U1B0As, with standard HCl is shown below: -What is the pH at point I (V₁/2 HCl added)?

Accepted Answer

At point I, half of the ascorbic acid has reacted with HCl to form the first equivalent of H2₂As, a diprotic acid. The buffer region has shifted to a higher pH due to the excess of Na₂As present. The pH at point I is around the pKa1 of ascorbic acid, which is 4.1. Since the pH is much higher than this value, both carboxylic acid groups are ionized, leaving H2₂As as the major species. The pH can be calculated using the Henderson-Hasselbalch equation: pH = pKa1 + log([Na₂As]/[H2₂As]) pH = 4.1 + log(1/2) pH = 4.1 + (-0.30) pH = 3.8 However, the addition of a strong acid like HCl will consume some of the Na₂As, shifting the equilibrium back towards H2₂As, and increasing the pH slightly. This can be estimated by assuming that the additional HCl forms H3O+ in solution, and calculating the new concentration of Na₂As: Na₂As + HCl → H2₂As + NaCl Initial [Na₂As] = 0.1 M At point I, [HCl] = 0.05 M [H3O+] = [HCl] = 0.05 M [H2₂As] = [Na₂As] - [HCl] = 0.05 M pH = pKa1 + log([Na₂As]/[H2₂As]) pH = 4.1 + log(0.05/0.05) pH = 4.1 + 0 pH = 4.1 This estimate is close to the true value, but the actual pH will be slightly higher due to the presence of other species in the mixture (e.g. Na+, H+, Ascorbate-). Looking at the titration curve, the pH at point I is around 11.8, which is closest to choice D.

Question 10

For carbonic acid (H₂CO₃), K_a1 = 4.30 *10^-⁷ and K_a2 = 5.62 *10^-¹¹. Calculate the pH of a 0.50 M solution of Na₂CO₃.

Accepted Answer

The correct answer is A because the pH of a 0.50 M solution of Na2CO3 can be calculated using the dissociation constants of carbonic acid (H2CO3) and the ICE table. The first dissociation constant (Ka1) for carbonic acid is 4.30 x 10^-7 and the second dissociation constant (Ka2) is 5.62 x 10^-11. The ICE table shows the initial concentrations of the species in the solution, the changes in concentrations after the first and second dissociation steps, and the equilibrium concentrations. The pH of the solution can then be calculated using the equilibrium concentrations of H+ and OH- ions.

Question 11

Calculate the pH when 200.0 mL of a 1.00 M solution of HS1U1B12S1U1B0A (KS1U1B1a1S1U1B0 = 1.0 * 10S1U1P1-S1S1P0S1U1P16S1S1P0, KS1U1B1a2S1U1B0 = 1.0 * 10S1U1P1-S1S1P0S1U1P110S1S1P0) is titrated with the following volumes of 1.00 M NaOH.
-0 mL of 1.00 M NaOH

Accepted Answer

The answer of Calculate the pH when 200.0 mL of...

Question 12

Calculate the pH when 200.0 mL of a 1.00 M solution of HS1U1B12S1U1B0A (KS1U1B1a1S1U1B0 = 1.0 * 10S1U1P1-S1S1P0S1U1P16S1S1P0, KS1U1B1a2S1U1B0 = 1.0 * 10S1U1P1-S1S1P0S1U1P110S1S1P0) is titrated with the following volumes of 1.00 M NaOH.
-200.0 mL of 1.00 M NaOH

Accepted Answer

The answer of Calculate the pH when 200.0 mL of...

Question 13

A 200.0-mL sample of the weak acid HS1U1B13S1U1B0A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.)
-After 300.0 mL of 0.200 M NaOH is added

Accepted Answer

The answer of A 200.0-mL sample of the weak acid...

Question 14

A 200.0-mL sample of the weak acid HS1U1B13S1U1B0A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.)
-After 100.0 mL of 0.200 M NaOH is added

Accepted Answer

The answer of A 200.0-mL sample of the weak acid...

Question 15

Calculate the pH when 200.0 mL of a 1.00 M solution of HS1U1B12S1U1B0A (KS1U1B1a1S1U1B0 = 1.0 * 10S1U1P1-S1S1P0S1U1P16S1S1P0, KS1U1B1a2S1U1B0 = 1.0 * 10S1U1P1-S1S1P0S1U1P110S1S1P0) is titrated with the following volumes of 1.00 M NaOH.
-250.0 mL of 1.00 M NaOH

Accepted Answer

First, we need to determine the stoichiometry of the reaction between H₂A and NaOH. From the formula of H₂A, we can see that it is a weak acid. Therefore, the reaction is: H₂A + NaOH → Na₂A + H2O The balanced equation shows a 1:1 mole ratio between H₂A and NaOH. This means that the number of moles of NaOH added to the solution is equal to the number of moles of H₂A initially present. We can use this fact to calculate the number of moles of H₂A remaining in the solution after each addition of NaOH. Before any NaOH is added: - The initial number of moles of H₂A is: 1.00 M × 0.2000 L = 0.2000 mol - The initial volume of the solution is: 0.2000 L - The initial pH can be calculated using the equation for a weak acid: pH = pKa + log([A-]/[HA]) where pKa is the acid dissociation constant of H₂A, [A-] is the concentration of the conjugate base of H₂A (i.e. ₂A-), and [HA] is the concentration of H₂A. From the formula, we can see that the conjugate base is formed when H₂A loses one proton (H+). Therefore, the Ka of H₂A is equal to the acidity constant of its conjugate base: Ka = Kw/Kb = 1.00 × 10^-14/1.00 × 10^-10 = 1.00 × 10^-4. Substituting Ka and the initial concentration of H₂A into the pH equation: pH = 4.00 + log(0.000200) = 3.70 After adding 250.0 mL of NaOH: - The total volume of the solution is: 0.2000 L + 0.2500 L = 0.4500 L - The number of moles of NaOH added is: 1.00 M × 0.2500 L = 0.2500 mol - The number of moles of H₂A remaining is: 0.2000 mol - 0.2500 mol = -0.0500 mol (this negative value indicates that all of the acid has been neutralized and excess NaOH is present) - The concentration of NaOH after adding 250.0 mL is: 0.2500 mol/0.4500 L = 0.5556 M - The OH- concentration after adding 250.0 mL is: 0.5556 M × 1.00 × 10^-14 = 5.56 × 10^-15 M - The pOH after adding 250.0 mL is: -log(5.56 × 10^-15) = 14.25 - The pH after adding 250.0 mL is: 14.00 - 14.25 = -0.25 (this negative value indicates that the solution is now basic) After adding an additional 250.0 mL of NaOH: - The total volume of the solution is: 0.2000 L + 0.5000 L = 0.7000 L - The number of moles of NaOH added is: 1.00 M × 0.2500 L = 0.2500 mol (for a total of 0.5000 mol added so far) - The number of moles of H₂A remaining is: 0.2000 mol - 0.5000 mol = -0.3000 mol (again, this negative value indicates excess NaOH) - The concentration of NaOH after adding 500.0 mL is: 0.5000 mol/0.7000 L = 0.7143 M - The OH- concentration after adding 500.0 mL is: 0.7143 M × 1.00 × 10^-14 = 7.14 × 10^-15 M - The pOH after adding 500.0 mL is: -log(7.14 × 10^-15) = 14.14 - The pH after adding 500.0 mL is: 14.00 - 14.14 = -0.14 (still basic, but less so than before) None of the answer choices match with the calculated pH after adding 500.0 mL of NaOH. However, we can see from the previous calculation that the pH is decreasing as more NaOH is added. This indicates that the pH after adding 250.0 mL of NaOH is likely to be higher than the pH after adding 500.0 mL of NaOH. Therefore, the best choice is B) 9.52.

Question 16

You dissolve 1.24 g of an unknown diprotic acid in 220.0 mL of H₂O. This solution is just neutralized by 6.44 mL of a 1.25 M NaOH solution. What is the molar mass of the unknown acid?

Accepted Answer

The diprotic acid reacts with NaOH in a 1:2 ratio. Calculate moles of NaOH: 6.44 mL * 1.25 M = 0.00805 mol. Moles of acid = 0.00805 mol / 2 = 0.004025 mol. Molar mass = 1.24 g / 0.004025 mol = 308 g/mol.

Question 17

Consider the following information about the diprotic acid ascorbic acid (HS1U1B12S1U1B0As for short, molar mass = 176.1).  The titration curve for disodium ascorbate, NaS1U1B12S1U1B0As, with standard HCl is shown below: 
-What is the pH at point III?

Accepted Answer

Point III corresponds to the second equivalence point, where both protons of ascorbic acid have been neutralized by the HCl titrant. At this point, the solution contains an excess of HCl, which dissociates completely to form H3O+ ions. The pH can be calculated using the equation for the second dissociation of ascorbic acid: H₂As- + H2O ⇌ ₂As2- + H3O+ The equilibrium expression for this reaction is: Ka2 = [₂As2-][H3O+] / [H₂As-] At the second equivalence point, [H₂As-] = 0 and [₂As2-] = [H3O+]. Substituting these values into the equilibrium expression and solving for pH, we get: Ka2 = (x)(x) / (C0 - x) = 1.6 × 10^-12 x = [H3O+] = 8.0 × 10^-5 pH = -log[H3O+] = 4.10

Question 18

Consider the titration of 200.0 mL of a 0.100 M solution of the weak acid H₂A with 0.200 M NaOH. The first equivalence point is reached after 100.0 mL of 0.200 M NaOH has been added, and the pH is 6.27. The pH after 65.0 mL of 0.200 M NaOH has been added, the pH is 4.95. Calculate the pH of the original acid (no NaOH added).

Accepted Answer

The answer of Consider the titration of 200.0 mL of...

Question 19

A solution contains 25 mmol of H₃PO₄ and 10. mmol of NaH₂PO₄. What volume of 2.0 M NaOH must be added to reach the second equivalence point of the titration of the H₃PO₄ with NaOH?

Accepted Answer

The answer of A solution contains 25 mmol of H₃PO₄...

Question 20

In the titration of 100.0 mL of a 0.200 M solution of H₂A (K_a1 = 1.0 * 10^-⁵, K_a2 = 1.0 * 10^-⁸), what volume of 0.400 M NaOH must be added to reach a pH of 5.00?

Accepted Answer

The first step is to write out the balanced chemical equation for the reaction: H₂A (acid) + NaOH (base) → Na₂A (salt) + H2O Next, we need to determine how many moles of acid are present in the solution: mol acid = Molarity x volume = 0.200 M x 0.100 L = 0.020 moles Since the acid is a weak acid, we can assume that at equilibrium, the amount of acid that has reacted with the base (NaOH) is negligible compared to the initial amount of acid. Therefore, we can assume that the amount of acid that remains in the solution is equal to the initial amount of acid (0.020 moles). To reach a pH of 5.00, we can assume that we have reached the equivalence point, where all of the acid has reacted with the base. At the equivalence point, the number of moles of acid is equal to the number of moles of base added: moles acid = moles base 0.020 moles acid = 0.400 M x volume base volume base = 0.020 moles acid / (0.400 M) = 0.050 L = 50.0 mL Therefore, the volume of 0.400 M NaOH needed to reach a pH of 5.00 is 50.0 mL.

Quiz 8: Applications of Aqueous Equilibria

A solution containing 10. mmol of CO₃²^- and 5.0 mmol of HCO₃^- is titrated with 1.0 M HCl. What total volume of HCl must be added to reach the second equivalence point?

A 200.0-mL sample of the weak acid H₃A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 220.0 mL of 0.200 M NaOH is added

A 0.210-g sample of an acid (molar mass = 192 g/mol) is titrated with 30.5 mL of 0.108 M NaOH to a phenolphthalein endpoint. The formula of the acid is

A solution contains 10. mmol of H₃PO₄ and 5.0 mmol of NaH₂PO₄. How many milliliters of 0.10 M NaOH must be added to reach the second equivalence point of the titration of the H₃PO₄ with NaOH?

Calculate the pH when 200.0 mL of a 1.00 M solution of H₂A (K_a1 = 1.0 * 10^-⁶, K_a2 = 1.0 * 10^-¹⁰) is titrated with the following volumes of 1.00 M NaOH. -150.0 mL of 1.00 M NaOH

A 5.95-g sample of an acid, H₂X, requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (removing both protons) . What is the molar mass of the acid expressed in grams per mol?

A 135.0-mL sample of a 0.25 M solution of H₃PO₄ is titrated with 0.12 M NaOH. What volume of base must be added to reach the third equivalence point?

A 200.0-mL sample of the weak acid H₃A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 75.0 mL of 0.200 M NaOH is added

Consider the following information about the diprotic acid ascorbic acid (H₂As for short, molar mass = 176.1) . The titration curve for disodium ascorbate, Na₂As, with standard HCl is shown below: -What is the pH at point I (V₁/2 HCl added) ?

For carbonic acid (H₂CO₃) , K_a1 = 4.30 10^-⁷ and K_a2 = 5.62 10^-¹¹. Calculate the pH of a 0.50 M solution of Na₂CO₃.

Calculate the pH when 200.0 mL of a 1.00 M solution of H₂A (K_a1 = 1.0 * 10^-⁶, K_a2 = 1.0 * 10^-¹⁰) is titrated with the following volumes of 1.00 M NaOH. -0 mL of 1.00 M NaOH

Calculate the pH when 200.0 mL of a 1.00 M solution of H₂A (K_a1 = 1.0 * 10^-⁶, K_a2 = 1.0 * 10^-¹⁰) is titrated with the following volumes of 1.00 M NaOH. -200.0 mL of 1.00 M NaOH

A 200.0-mL sample of the weak acid H₃A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 300.0 mL of 0.200 M NaOH is added

A 200.0-mL sample of the weak acid H₃A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 100.0 mL of 0.200 M NaOH is added

Calculate the pH when 200.0 mL of a 1.00 M solution of H₂A (K_a1 = 1.0 * 10^-⁶, K_a2 = 1.0 * 10^-¹⁰) is titrated with the following volumes of 1.00 M NaOH. -250.0 mL of 1.00 M NaOH

You dissolve 1.24 g of an unknown diprotic acid in 220.0 mL of H₂O. This solution is just neutralized by 6.44 mL of a 1.25 M NaOH solution. What is the molar mass of the unknown acid?

Consider the following information about the diprotic acid ascorbic acid (H₂As for short, molar mass = 176.1) . The titration curve for disodium ascorbate, Na₂As, with standard HCl is shown below: -What is the pH at point III?

A solution contains 25 mmol of H₃PO₄ and 10. mmol of NaH₂PO₄. What volume of 2.0 M NaOH must be added to reach the second equivalence point of the titration of the H₃PO₄ with NaOH?

In the titration of 100.0 mL of a 0.200 M solution of H₂A (K_a1 = 1.0 * 10^-⁵, K_a2 = 1.0 * 10^-⁸) , what volume of 0.400 M NaOH must be added to reach a pH of 5.00?

Quiz 8: Applications of Aqueous Equilibria

A solution containing 10. mmol of CO32- and 5.0 mmol of HCO3- is titrated with 1.0 M HCl. What total volume of HCl must be added to reach the second equivalence point?

A 200.0-mL sample of the weak acid H3A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 220.0 mL of 0.200 M NaOH is added

A 0.210-g sample of an acid (molar mass = 192 g/mol) is titrated with 30.5 mL of 0.108 M NaOH to a phenolphthalein endpoint. The formula of the acid is

A solution contains 10. mmol of H3PO4 and 5.0 mmol of NaH2PO4. How many milliliters of 0.10 M NaOH must be added to reach the second equivalence point of the titration of the H3PO4 with NaOH?

Calculate the pH when 200.0 mL of a 1.00 M solution of H2A (Ka1 = 1.0 * 10-6, Ka2 = 1.0 * 10-10) is titrated with the following volumes of 1.00 M NaOH. -150.0 mL of 1.00 M NaOH

A 5.95-g sample of an acid, H2X, requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (removing both protons) . What is the molar mass of the acid expressed in grams per mol?

A 135.0-mL sample of a 0.25 M solution of H3PO4 is titrated with 0.12 M NaOH. What volume of base must be added to reach the third equivalence point?

A 200.0-mL sample of the weak acid H3A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 75.0 mL of 0.200 M NaOH is added

Consider the following information about the diprotic acid ascorbic acid (H2As for short, molar mass = 176.1) . The titration curve for disodium ascorbate, Na2As, with standard HCl is shown below: -What is the pH at point I (V1/2 HCl added) ?

For carbonic acid (H2CO3) , Ka1 = 4.30 *10-7 and Ka2 = 5.62 *10-11. Calculate the pH of a 0.50 M solution of Na2CO3.

Calculate the pH when 200.0 mL of a 1.00 M solution of H2A (Ka1 = 1.0 * 10-6, Ka2 = 1.0 * 10-10) is titrated with the following volumes of 1.00 M NaOH. -0 mL of 1.00 M NaOH

Calculate the pH when 200.0 mL of a 1.00 M solution of H2A (Ka1 = 1.0 * 10-6, Ka2 = 1.0 * 10-10) is titrated with the following volumes of 1.00 M NaOH. -200.0 mL of 1.00 M NaOH

A 200.0-mL sample of the weak acid H3A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 300.0 mL of 0.200 M NaOH is added

A 200.0-mL sample of the weak acid H3A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 100.0 mL of 0.200 M NaOH is added

Calculate the pH when 200.0 mL of a 1.00 M solution of H2A (Ka1 = 1.0 * 10-6, Ka2 = 1.0 * 10-10) is titrated with the following volumes of 1.00 M NaOH. -250.0 mL of 1.00 M NaOH

You dissolve 1.24 g of an unknown diprotic acid in 220.0 mL of H2O. This solution is just neutralized by 6.44 mL of a 1.25 M NaOH solution. What is the molar mass of the unknown acid?

Consider the following information about the diprotic acid ascorbic acid (H2As for short, molar mass = 176.1) . The titration curve for disodium ascorbate, Na2As, with standard HCl is shown below: -What is the pH at point III?

A solution contains 25 mmol of H3PO4 and 10. mmol of NaH2PO4. What volume of 2.0 M NaOH must be added to reach the second equivalence point of the titration of the H3PO4 with NaOH?

In the titration of 100.0 mL of a 0.200 M solution of H2A (Ka1 = 1.0 * 10-5, Ka2 = 1.0 * 10-8) , what volume of 0.400 M NaOH must be added to reach a pH of 5.00?

A solution containing 10. mmol of CO₃²^- and 5.0 mmol of HCO₃^- is titrated with 1.0 M HCl. What total volume of HCl must be added to reach the second equivalence point?

A 200.0-mL sample of the weak acid H₃A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 220.0 mL of 0.200 M NaOH is added

A solution contains 10. mmol of H₃PO₄ and 5.0 mmol of NaH₂PO₄. How many milliliters of 0.10 M NaOH must be added to reach the second equivalence point of the titration of the H₃PO₄ with NaOH?

Calculate the pH when 200.0 mL of a 1.00 M solution of H₂A (K_a1 = 1.0 * 10^-⁶, K_a2 = 1.0 * 10^-¹⁰) is titrated with the following volumes of 1.00 M NaOH. -150.0 mL of 1.00 M NaOH

A 5.95-g sample of an acid, H₂X, requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (removing both protons) . What is the molar mass of the acid expressed in grams per mol?

A 135.0-mL sample of a 0.25 M solution of H₃PO₄ is titrated with 0.12 M NaOH. What volume of base must be added to reach the third equivalence point?

A 200.0-mL sample of the weak acid H₃A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 75.0 mL of 0.200 M NaOH is added

Consider the following information about the diprotic acid ascorbic acid (H₂As for short, molar mass = 176.1) . The titration curve for disodium ascorbate, Na₂As, with standard HCl is shown below: -What is the pH at point I (V₁/2 HCl added) ?

For carbonic acid (H₂CO₃) , K_a1 = 4.30 10^-⁷ and K_a2 = 5.62 10^-¹¹. Calculate the pH of a 0.50 M solution of Na₂CO₃.

Calculate the pH when 200.0 mL of a 1.00 M solution of H₂A (K_a1 = 1.0 * 10^-⁶, K_a2 = 1.0 * 10^-¹⁰) is titrated with the following volumes of 1.00 M NaOH. -0 mL of 1.00 M NaOH

Calculate the pH when 200.0 mL of a 1.00 M solution of H₂A (K_a1 = 1.0 * 10^-⁶, K_a2 = 1.0 * 10^-¹⁰) is titrated with the following volumes of 1.00 M NaOH. -200.0 mL of 1.00 M NaOH

A 200.0-mL sample of the weak acid H₃A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 300.0 mL of 0.200 M NaOH is added

A 200.0-mL sample of the weak acid H₃A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.) -After 100.0 mL of 0.200 M NaOH is added

Calculate the pH when 200.0 mL of a 1.00 M solution of H₂A (K_a1 = 1.0 * 10^-⁶, K_a2 = 1.0 * 10^-¹⁰) is titrated with the following volumes of 1.00 M NaOH. -250.0 mL of 1.00 M NaOH

You dissolve 1.24 g of an unknown diprotic acid in 220.0 mL of H₂O. This solution is just neutralized by 6.44 mL of a 1.25 M NaOH solution. What is the molar mass of the unknown acid?

Consider the following information about the diprotic acid ascorbic acid (H₂As for short, molar mass = 176.1) . The titration curve for disodium ascorbate, Na₂As, with standard HCl is shown below: -What is the pH at point III?

A solution contains 25 mmol of H₃PO₄ and 10. mmol of NaH₂PO₄. What volume of 2.0 M NaOH must be added to reach the second equivalence point of the titration of the H₃PO₄ with NaOH?

In the titration of 100.0 mL of a 0.200 M solution of H₂A (K_a1 = 1.0 * 10^-⁵, K_a2 = 1.0 * 10^-⁸) , what volume of 0.400 M NaOH must be added to reach a pH of 5.00?