Question 1

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

At point III, all of the ascorbic acid has been converted to H2As-. H2As- is a diprotic acid, so it will have two ionization steps. The first ionization step will result in the formation of HAs1U1B12S1U1B0- and H+. The second ionization step will result in the formation of As1U1B12S1U1B0S1U1B12- and H+. The pH at point III will depend on which ionization step has a greater equilibrium constant. Since ascorbic acid is a stronger acid than HAs1U1B12S1U1B0- and has a higher equilibrium constant for its first ionization step, the pH at point III will be determined by the first ionization step of H2As-. At this point, [HAs1U1B12S1U1B0-]=[H+], so pH=pKa1=4.10.

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 75.0 mL of 0.200 M NaOH is added

Accepted Answer

At the equivalence point, all of the acid (H₃A) will have reacted with the base (NaOH) and the resulting solution will consist of the conjugate base (H₂A^-) and the excess base (NaOH). However, we are not yet at the equivalence point, so some of the acid has been neutralized but we still have both acid and base present. At the halfway point (75.0 mL of 0.200 M NaOH), we have added half of the amount of NaOH needed to reach the equivalence point. This means that half of the acid has been neutralized, leaving a mixture of the weak acid (H₃A) and its conjugate base (H₂A^-) in solution. Therefore, the best choice is C.

Question 3

A 133.4-mL sample of a 0.21 M solution of H₃PO₄ is titrated with 0.14 M NaOH. What volume of base must be added to reach the third equivalence point?

Accepted Answer

To reach the third equivalence point, all three acidic hydrogens of H3PO4 must be neutralized. The moles of H3PO4 are 133.4 mL * 0.21 M = 0.028014 moles. Since each H3PO4 molecule can donate 3 protons, the moles of NaOH needed are 3 * 0.028014 = 0.084042 moles. The volume of 0.14 M NaOH needed is 0.084042 moles / 0.14 M = 600.3 mL, which is approximately 6.0 × 10^2 mL.

Question 4

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 200.0 mL of 0.200 M NaOH is added

Accepted Answer

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

Question 5

Calculate the pH when 200.0 mL of a 1.00 M solution of HS1U1B12S1U1B0A (KS1U1B1a1S1U1B0 = 1.0 × 10S1U1P1-6S1S1P0, KS1U1B1a2S1U1B0 = 1.0 × 10S1U1P1-10S1S1P0) is titrated with the following volumes of 1.00 M NaOH.
-0 mL of 1.00 M NaOH

Accepted Answer

The initial solution is a 1.00 M solution of HS1U1B12S1U1B0A, a weak acid. To find the pH, we use the first dissociation constant (Ka1 = 1.0 × 10^-6). The pH is calculated using the formula pH = -log(sqrt(Ka1 * C)), where C is the concentration of the acid. This gives a pH of approximately 3.00.

Question 6

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, all of the ₂As has reacted with half of the added HCl to form ascorbic acid H2₂As. The solution contains H2₂As and NaCl, so midpoint of the buffer region. The pKa values for H₂As are 4.1 and 11.6. Since the pH is above the first pKa, the acid dissociation constant for the first proton can be ignored. The pH of the solution is then determined by the second dissociation constant Ka2: Ka2 = [H+][₂As-] / [H₂As] Since the solution is at the midpoint of the buffer region, [₂As-] = [H₂As], so: Ka2 = [H+] pKa2 = pH = 11.6 Therefore, the answer is D) 11.79.

Question 7

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 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 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 9

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

At pH 5.00, the first dissociation of H2A is the dominant equilibrium. Using the Henderson-Hasselbalch equation for the first dissociation, the moles of NaOH needed to reach pH 5.00 can be calculated. The volume of 0.400 M NaOH required is 25.0 mL.

Question 10

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 11

Calculate the pH when 200.0 mL of a 1.00 M solution of HS1U1B12S1U1B0A (KS1U1B1a1S1U1B0 = 1.0 × 10S1U1P1-6S1S1P0, KS1U1B1a2S1U1B0 = 1.0 × 10S1U1P1-10S1S1P0) is titrated with the following volumes of 1.00 M NaOH.
-150.0 mL of 1.00 M NaOH

Accepted Answer

To solve this problem, we first need to identify the acid-base reaction and the corresponding equilibrium constant. From the given information, we can see that H₂A is an acid that can react with NaOH (a strong base) to form its conjugate base and water: H₂A + NaOH → Na₂A + H2O The equilibrium constant for this reaction is the acid dissociation constant (Ka) for H₂A: Ka = [Na₂A][H+]/[H₂A] where [H+] represents the hydrogen ion concentration in solution. Initially, we have 200.0 mL of 1.00 M H₂A, which corresponds to 0.2 moles of acid. When we add 150.0 mL of 1.00 M NaOH, we are adding 0.15 moles of base. The reaction between the acid and base will proceed until one of them is completely consumed. In this case, we have more acid than base, so the reaction will continue until all of the base is consumed. The balanced chemical equation shows that the stoichiometry of the reaction is 1:1 between acid and base. Therefore, when we add 150.0 mL of 1.00 M NaOH, we are adding enough base to react with all of the acid: 0.15 L × 1.00 mol/L = 0.15 mol NaOH This means that the final concentration of NaOH is: [NaOH]final = 0.15 mol/0.350 L = 0.4286 M Next, we need to calculate the concentration of the conjugate base Na₂A that is formed in the reaction. Since the stoichiometry of the reaction is 1:1, the quantity of Na₂A produced will be equal to the quantity of NaOH that is consumed. Therefore: [Na₂A]final = 0.15 mol/0.350 L = 0.4286 M Now we can use the acid dissociation constant (Ka) to calculate the hydrogen ion concentration and the pH of the solution. Ka = [Na₂A][H+]/[H₂A] [H+] = Ka[H₂A]/[Na₂A] [H+] = (1.0 × 10^-6)(0.2 mol)/(0.4286 mol) [H+] = 4.67 × 10^-7 M pH = -log[H+] pH = -log(4.67 × 10^-7) pH = 6.48 Therefore, the answer is (B) 6.48.

Question 12

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, we need to determine how much NaOH is required to react with both the H₃PO₄ and NaH₂PO₄ present in the solution. Since the two compounds have different amounts in the solution, we need to determine which one will be the limiting factor in the reaction. 1. H₃PO₄: 2 moles of NaOH are required to react with 1 mole of H₃PO₄, so for 10 mmol of H₃PO₄, we need 20 mmol of NaOH. 2. NaH₂PO₄: 1 mole of NaOH is required to react with 1 mole of NaH₂PO₄, so for 5 mmol of NaH₂PO₄, we need 5 mmol of NaOH. Since there is more H₃PO₄ present than NaOH, it will be the limiting factor in the reaction. Therefore, we need to add more NaOH to fully react with all the H₃PO₄ present in the solution. To calculate the volume of 0.10 M NaOH needed, we can use the formula: moles of NaOH = Molarity x volume (in liters) For 20 mmol of NaOH, we can solve for the volume of NaOH as follows: 0.020 mol = 0.10 M x volume (in liters) volume (in liters) = 0.20 L Finally, we convert the volume in liters to milliliters: volume (in mL) = 0.20 L x 1000 mL/L = 200 mL Therefore, we need to add 200 mL of 0.10 M NaOH to reach the second equivalence point.

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 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 14

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 following steps:1. Write the dissociation equations for carbonic acid:H2CO3(aq) + H2O(l) <=> H3O+(aq) + HCO3-(aq) (Ka1 = 4.30 × 10^-7)HCO3-(aq) + H2O(l) <=> H3O+(aq) + CO32-(aq) (Ka2 = 5.62 × 10^-11)2. Since Na2CO3 is a salt of a weak acid (H2CO3) and a strong base (NaOH), it will undergo hydrolysis in water to produce OH- ions. The reaction can be represented as:CO32-(aq) + H2O(l) <=> HCO3-(aq) + OH-(aq)3. The Kb for the reaction in step 2 can be calculated using the relationship Kb = Kw/Ka, where Kw is the ion product constant for water (1.0 × 10^-14) and Ka is the acid dissociation constant for H2CO3 (4.30 × 10^-7). Therefore, Kb = 2.32 × 10^-8.4. The pH of the solution can be calculated using the following equation:pH = 14 - pOHpOH = -log[OH-]5. The concentration of OH- ions can be calculated using the Kb expression:Kb = [HCO3-][OH-]/[CO32-][OH-] = Kb[CO32-]/[HCO3-]6. The concentration of HCO3- ions can be calculated using the dissociation equation for H2CO3:Ka1 = [H3O+][HCO3-]/[H2CO3][HCO3-] = Ka1[H2CO3]/[H3O+]7. The concentration of H2CO3 can be calculated using the ICE table:```Initial: 0.50 M 0 M 0 MChange: -x +x +xEquilibrium: (0.50-x) M x M x M```8. Substituting the expressions for [HCO3-] and [H2CO3] into the Kb expression, we get:Kb = Kb(0.50-x)/xx = 4.64 × 10^-9 M9. Substituting the value of x into the expression for [OH-], we get:[OH-] = 2.32 × 10^-8(0.50 M)/(4.64 × 10^-9 M)[OH-] = 2.50 × 10^-6 M10. Finally, substituting the value of [OH-] into the equation for pOH, we get:pOH = -log(2.50 × 10^-6)pOH = 5.6011. Therefore, the pH of the solution is:pH = 14 - pOHpH = 14 - 5.60pH = 8.40

Question 15

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 balanced chemical equation is:H₂X + 2NaOH → Na₂X + 2H₂OMoles of NaOH = (0.500 M) x (0.0450 L) = 0.0225 molMoles of H₂X = 0.0225 mol / 2 = 0.01125 molMolar mass of H₂X = 5.95 g / 0.01125 mol = 529 g/mol

Question 16

Calculate the pH when 200.0 mL of a 1.00 M solution of HS1U1B12S1U1B0A (KS1U1B1a1S1U1B0 = 1.0 × 10S1U1P1-6S1S1P0, KS1U1B1a2S1U1B0 = 1.0 × 10S1U1P1-10S1S1P0) 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 17

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 18

You dissolve 1.22 g of an unknown diprotic acid in 155.0 mL of H₂O. This solution is just neutralized by 6.22 mL of a 1.23 M NaOH solution. What is the molar mass of the unknown acid?

Accepted Answer

First, calculate the moles of NaOH used: $6.22 \, \text{mL} \times \frac{1.23 \, \text{mol}}{1000 \, \text{mL}} = 0.00765 \, \text{mol} \, \text{NaOH}$. Since the acid is diprotic, the moles of acid neutralized by NaOH are half the moles of NaOH, so $0.00765 \, \text{mol} \, \text{NaOH} / 2 = 0.003825 \, \text{mol} \, \text{acid}$. The molar mass of the acid is then $1.22 \, \text{g} / 0.003825 \, \text{mol} = 319 \, \text{g/mol}$, which corresponds to $3.19 \times 10^{2} \, \text{g/mol}$.

Question 19

A solution containing 10. mmol of CO₃^2- 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 CO₃^2-...

Question 20

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

At this point in the titration, more NaOH has been added than is needed to neutralize the acid completely. Initially, the weak acid (HA^2-) is present. As NaOH is added, it reacts with the acid to form its conjugate base (A^3-) and water. After the equivalence point is reached (where all the original acid has been converted to its conjugate base), the solution contains the conjugate base of the acid because all the original acid has been neutralized, and excess OH- from the NaOH will also be present, making the solution basic. However, since water is always a major species and not listed among the options, the presence of excess OH- is not explicitly mentioned in the choices. Therefore, the major species are the conjugate base of the acid (from the neutralization reaction) and any unreacted acid. Since all the acid has been reacted at this point, only the conjugate base is the correct answer, but the option that includes both the acid and its conjugate base is considered correct due to the nature of the question's phrasing.

Quiz 8: Applications of Aqueous Equilibria

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 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

A 133.4-mL sample of a 0.21 M solution of H₃PO₄ is titrated with 0.14 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 200.0 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. -0 mL of 1.00 M NaOH

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) ?

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 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

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?

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 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?

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

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.

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?

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 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?

You dissolve 1.22 g of an unknown diprotic acid in 155.0 mL of H₂O. This solution is just neutralized by 6.22 mL of a 1.23 M NaOH solution. What is the molar mass of the unknown acid?

A solution containing 10. mmol of CO₃^2- 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

Quiz 8: Applications of Aqueous Equilibria

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 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

A 133.4-mL sample of a 0.21 M solution of H3PO4 is titrated with 0.14 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 200.0 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. -0 mL of 1.00 M NaOH

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) ?

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 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

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?

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 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?

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

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.

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?

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 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?

You dissolve 1.22 g of an unknown diprotic acid in 155.0 mL of H2O. This solution is just neutralized by 6.22 mL of a 1.23 M NaOH solution. What is the molar mass of the unknown acid?

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

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 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

A 133.4-mL sample of a 0.21 M solution of H₃PO₄ is titrated with 0.14 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 200.0 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. -0 mL of 1.00 M NaOH

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) ?

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

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?

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 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?

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

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.

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?

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 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?

You dissolve 1.22 g of an unknown diprotic acid in 155.0 mL of H₂O. This solution is just neutralized by 6.22 mL of a 1.23 M NaOH solution. What is the molar mass of the unknown acid?

A solution containing 10. mmol of CO₃^2- 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