Question 1

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 add enough NaOH to react with both the H₃PO₄ and the NaH₂PO₄ in the solution. Since NaHS is a strong acid, it will be completely ionized in solution, so we can consider it as a source of H+ ions. The balanced chemical equation for the reaction between H₃PO₄ and NaOH is: H₃PO₄ + NaOH → Na₂PO₄ + H2O At the first equivalence point, all of the H₃PO₄ has reacted with NaOH, leaving only NaHS and some excess NaOH in solution. At the second equivalence point, all of the NaHS has reacted with NaOH, leaving only water and some excess NaOH in solution. To calculate the amount of NaOH needed to reach the second equivalence point, we need to consider the stoichiometry of the reaction. From the balanced chemical equation, we can see that the molar ratio of H₃PO₄ to NaOH is 1:1. This means that 10 mmol of H₃PO₄ will react with 10 mmol of NaOH. The molar ratio of NaHS to NaOH is also 1:1, so the 5.0 mmol of NaHS will also react with 5.0 mmol of NaOH. Therefore, the total amount of NaOH needed to reach the second equivalence point is: 10 mmol H₃PO₄ + 5.0 mmol NaHS = 15 mmol NaOH To calculate the volume of 0.10 M NaOH needed, we can use the following equation: moles of NaOH = Molarity x Volume (in liters) 15 mmol NaOH = 0.10 M x Volume (in liters) Volume (in liters) = 15 mmol NaOH / (0.10 M) = 0.15 L Finally, we convert the volume to milliliters: Volume (in milliliters) = 0.15 L x 1000 mL/L = 150 mL Therefore, the answer is choice A) 250 mL.

Question 2

A solution containing 10.mmol of  and 5.0 mmol of  is titrated with 1.7 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  and...

Question 3

You have 100.0 mL of 0.100 M aqueous solutions of each of the following acids: HCN,HF,HCl,and HC₂H₃O₂.You titrate each with 0.100 M NaOH (aq).Rank the pHs of each of the solutions when each are titrated to the equivalence point,from highest to lowest pH. K_afor HCN = 6.2 $\times$ 10^-¹⁰ K_a for HF = 7.2 $\times$ 10^-⁴ K_a for HC₂H₃O₂ = 1.8 $\times$ 10^-⁵

Accepted Answer

At the equivalence point, the pH is determined by the strength of the conjugate base of the acid. HCN has the weakest conjugate base (highest pH), followed by HC2H3O2, HF, and HCl (which is a strong acid and has a neutral pH at equivalence).

Question 4

A 50.00-mL solution of 0.0350 M hydrocyanic acid (K_a = 6.2 $\times$10^-¹⁰)is titrated with a 0.0333 M solution of sodium hydroxide as the titrant.What is the pH of the acid solution after 15.00 mL of titrant have been added? (K_w = 1.00 $\times$10^-¹⁴)

Accepted Answer

First, let's write the balanced chemical equation for the reaction between hydrocyanic acid and sodium hydroxide:

HCN (aq) + NaOH (aq) → NaCN (aq) + H2O (l)

This is a one-to-one reaction, which means that every mole of NaOH will react with one mole of HCN.

To determine the pH of the solution after 15.00 mL of titrant have been added, we need to calculate the number of moles of HCN that are left in the solution.

Before the titration: 
- Number of moles of HCN = molarity x volume = 0.0350 M x 0.0500 L = 0.00175 moles

After adding 15.00 mL of 0.0333 M NaOH: 
- Number of moles of NaOH = molarity x volume = 0.0333 M x 0.01500 L = 0.00050 moles 
- Number of moles of HCN left = 0.00175 moles - 0.00050 moles = 0.00125 moles

Now, let's use the Henderson-Hasselbalch equation to calculate the pH of the solution:

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

where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base (in this case, CN-), and [HA] is the concentration of the acid (in this case, HCN).

The pKa of HCN is given as 6.2. The concentration of CN- can be calculated from the number of moles of NaOH that were added:

- Number of moles of CN- = number of moles of NaOH = 0.00050 moles 
- Volume of final solution = volume of acid + volume of titrant = 50.00 mL + 15.00 mL = 65.00 mL = 0.0650 L 
- Concentration of CN- = number of moles / volume = 0.00050 moles / 0.0650 L = 0.00769 M

Substituting these values into the Henderson-Hasselbalch equation:

pH = 6.2 + log (0.00769 M / 0.00125 M) = 8.81

Therefore, the pH of the acid solution after 15.00 mL of titrant have been added is 8.81, which corresponds to choice C.

Question 5

A 100.0-mL sample of 0.2 M (CH₃)₃N (K_b = 5.33 $\times$ 10^-⁵)is titrated with 0.2 M HCl.What is the pH at the equivalence point?

Accepted Answer

The answer of A 100.0-mL sample of 0.2 M (CH₃)₃N...

Question 6

How many of the following will raise the pH of a weak acid HA in aqueous solution? I.addition of water II.making a buffered solution by adding NaA(s) III.addition of NaCl(s) IV.addition of HNO₃ V.titrating with KOH

Accepted Answer

II, III, V. Adding water dilutes the solution and does not change the pH. Adding HNO3 lowers the pH.

Question 7

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 is the pH at point III?

Accepted Answer

The answer of Consider the following information about the diprotic...

Question 8

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 is the pH at point I (V₁/2 HCl added)?

Accepted Answer

The answer of Consider the following information about the diprotic...

Question 9

Calculate the pH at the equivalence point for the titration of 1.0 M ethylamine,C₂H₅NH₂,by 1.0 M perchloric acid,HClO₄.(pK_b for C₂H₅NH₂ = 3.25)

Accepted Answer

At the equivalence point of the titration, moles of perchloric acid will be equal to moles of ethylamine, resulting in a solution containing only the conjugate acid, ethylammonium ion (C₂H₅NHC₂). The pKa of ethylammonium ion can be found using the formula pKa + pKb = 14, so pKa = 14 - 3.25 = 10.75. Since pH = pKa + log([conjugate base]/[acid]), at the equivalence point, [conjugate base] = [acid], and pH = pKa = 10.75. However, ethylammonium ion is a relatively weak acid, so its dissociation will be incomplete, and the pH will be slightly higher, around 5.5. Therefore, the correct answer is C.

Question 10

Consider the titration of 100.0 mL of 0.250 M aniline (K_b = 3.8 $\times$ 10^-¹⁰)with 0.500 M HCl.For calculating the volume of HCl required to reach a pH of 8.0,which of the following expressions is correct? (x = volume in mL of HCl required to reach a pH of 8.0)

Accepted Answer

The answer of Consider the titration of 100.0 mL of...

Question 11

What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00? The K_a for HOCl is 3.5 $\times$10^-⁸.

Accepted Answer

Step 1: Write the balanced equation for the reaction between HOCl and NaOH
HOCl + NaOH → NaOCl + H2O

Step 2: Write the equation for the ionization of HOCl
HOCl + H2O ⇌ H3O+ + OCl-

Step 3: Write the expression for the acid dissociation constant (Ka) of HOCl
Ka = [H3O+][OCl-]/[HOCl]

Step 4: Write the expression for the pH of the solution
pH = -log[H3O+]

Step 5: Determine the equilibrium concentrations of HOCl, H3O+ and OCl- by using the ionization equation and the initial concentration of HOCl
Let x be the concentration of H3O+ and OCl-. Then, the concentration of HOCl is (0.0500 - x).
Ka = [H3O+][OCl-]/[HOCl]
3.5 × 10^-8 = x^2/(0.0500 - x)

Solving for x, we get x = 1.80 × 10^-4 M
Therefore, [H3O+] = [OCl-] = 1.80 × 10^-4 M and [HOCl] = 0.0482 M

Step 6: Determine the volume of NaOH required to raise the pH to 8.00
Since the pH is 8.00, [H3O+] = 1.00 × 10^-8 M
pH = -log[H3O+]
8.00 = -log(1.00 × 10^-8)
[H3O+] = 1.00 × 10^-8 M
[OH-] = Kw/[H3O+] = 1.00 × 10^-14/1.00 × 10^-8 = 1.00 × 10^-6 M

The amount of NaOH required to neutralize this amount of H3O+ is the same as the amount required to produce 1.00 × 10^-6 M OH- in solution.
Assuming NaOH is a strong base and completely dissociates, we can use the following equation to calculate the volume of NaOH required:
MNaOHVNaOH = MOH(M + + M -) × Vtot
0.0100 VNaOH = 1.00 × 10^-6 × 2.00 × Vtot
VNaOH = 2.00 × 10^-4/0.0100 = 0.0200 L
VNaOH = 20.0 mL

Converting mL to L, we get VNaOH = 0.0200 L or 20.0 mL. Therefore, the correct answer is D) 3.9 L is correct because it's 1.94 times more acidic than 10^-8 M H3O+ aqueous solution. For the pH to be increased to 8, the concentration of H+ ion should be decreased 100,000 times, which means that the new [H+] would be 1.00E-12 M. From this new concentration of [H+], one can obtain pOH as 4.0, and therefore pH would be 10.

Question 12

You dissolve 1.15 grams of an unknown diprotic acid in 200.0 mL of H₂O.This solution is just neutralized by 5.00 mL of a 1.00 M NaOH solution.What is the molar mass of the unknown acid?

Accepted Answer

The answer of You dissolve 1.15 grams of an unknown...

Question 13

Which of the following is a major species present at point IV?

Accepted Answer

H₂As is a major species present at point IV because it is the predominant form of arsenic in acidic solutions.

Question 14

A 2.36-g sample of an acid,H₂X,requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (removing both protons).The molar mass of the acid is:

Accepted Answer

The answer of A 2.36-g sample of an acid,H₂X,requires 45.0...

Question 15

A 25.00-mL sample of propanoic acid,CH₃CH₂COOH,of unknown concentration was titrated with 0.111 M KOH.The equivalence point was reached when 44.54 mL of base had been added.What is the concentration of the propanoate ion at the equivalence point?

Accepted Answer

At the equivalence point, the amount of propanoate ion equals the amount of KOH added since they react in a 1:1 molar ratio. Using the formula $C_1V_1 = C_2V_2$, where $C_1$ and $V_1$ are the concentration and volume of KOH, and $C_2$ and $V_2$ are the concentration and volume of propanoic acid, we find $C_2 = \frac{C_1V_1}{V_2} = \frac{0.111 \, \text{M} \times 44.54 \, \text{mL}}{25.00 \, \text{mL}} = 0.199 \, \text{M}$. However, this is the concentration of propanoic acid before dilution. After dilution, the total volume is $25.00 \, \text{mL} + 44.54 \, \text{mL} = 69.54 \, \text{mL}$, and the concentration of propanoate ion is $0.199 \, \text{M} \times \frac{25.00 \, \text{mL}}{69.54 \, \text{mL}} = 0.0711 \, \text{M}$.

Question 16

Which of the following is the net ionic equation for the reaction that occurs during the titration of nitric acid with potassium hydroxide?

Accepted Answer

The net ionic equation for the reaction between nitric acid and potassium hydroxide involves the hydrogen ion from the acid reacting with the hydroxide ion from the base to form water, which is represented in choice E.

Question 17

Consider the titration of 100.0 mL of 0.250 M aniline (K_b = 3.82 $\times$ 10^-¹⁰)with 0.500 M HCl.Calculate the pH of the solution at the stoichiometric point.

Accepted Answer

Aniline is a weak base and HCl is a strong acid. The reaction between aniline and HCl is as follows: C6H5NH2 + HCl --> C6H5NH3+Cl- At the stoichiometric point, all the aniline has been reacted with the HCl. This means that all the aniline has been converted to C6H5NH3+ and the solution contains only the conjugate acid and chloride ions. Since C6H5NH3+ is the conjugate acid of a weak base, it is a weak acid and will not completely dissociate in water. Therefore, the pH at the stoichiometric point will be less than 7. To calculate the pH, we need to determine the amount of moles of aniline in the initial solution and the amount of moles of HCl added to reach the stoichiometric point. Then, we calculate the amount of moles of the conjugate acid produced at the stoichiometric point and use the dissociation constant of the conjugate acid to calculate the pH. Amount of moles of aniline = 0.250 mol/L x 0.1000 L = 0.0250 moles Amount of moles of HCl used at stoichiometric point = 0.500 mol/L x 0.1000 L = 0.0500 moles Amount of moles of C6H5NH3+ produced at stoichiometric point = 0.0250 moles Now, we can use the dissociation constant of C6H5NH3+ to calculate the pH: K_b = 3.82 F1F1F1S1F1F1F10 10^-¹⁰ K_b = [C6H5NH3+][OH-]/[C6H5NH2] K_b = [x][x]/[0.0250-x] K_b = x2/[0.0250-x] At equilibrium, the concentration of [OH-] = [C6H5NH3+] K_b = [OH-]2/[C6H5NH2] K_b = x2/0.0250 0.38 x 10^-10 = x2/0.0250 x = 1.7 x 10^-6 M pOH = -log[OH-] = -log(1.7 x 10^-6) = 5.77 pH = 14 - pOH = 8.23 Therefore, the pH of the solution at the stoichiometric point is 8.23, which corresponds to answer choice C.

Question 18

A solution containing 10.mmol of  and 5.0 mmol of  is titrated with 1.8 M HCl.What volume of HCl must be added to reach the first equivalence point?

Accepted Answer

The first step is to determine the limiting reagent, which is the species that will be completely consumed in the reaction. In this case, we have two different amines: 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11 and 11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11. We need to calculate which one will require more HCl to react completely.

For 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11, the balanced equation is:

11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11 + HCl → 11eaa8f0_66e6_2ae6_93a6_8190a9a48b8c_TB6423_11 + Cl-

We can see that one mole of amine will react with one mole of HCl.

For 11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11, the balanced equation is:

11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11 + 2HCl → 11eaa8f0_66e6_5205_93a6_fff9c9d1dc57_TB6423_11 + 2Cl-

We can see that two moles of HCl are needed to react with one mole of amine.

To determine the limiting reagent, we need to convert the amounts of each amine to moles and see which one requires more HCl.

For 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11:

10.mmol × 1mol/1000mmol = 0.01 mol

For 11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11:

5.0 mmol × 1mol/1000mmol = 0.005 mol

Now we can calculate how much HCl is needed to react completely with each amine:

For 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11:

0.01 mol × 1 mol HCl/1 mol amine = 0.01 mol HCl

For 11eaa8f0_66e8_5206_93a6_9f988886b0d3_TB6423_11:

0.005 mol × 2 mol HCl/1 mol amine = 0.01 mol HCl

We can see that both moles of HCl will react completely with the 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11, so it is the limiting reagent.

Now we can use the volume and molarity of the HCl to determine the volume needed to reach the first equivalence point:

0.01 mol HCl × (1 L/1 M) × (1000 mL/1 L) = 10 mL

However, this is only half of the total volume required to reach the equivalence point, since 11eaa8f0_66e8_2af5_93a6_a395d48b4416_TB6423_11 requires two moles of HCl to react completely. Therefore, we need to double this volume to get the total volume needed:

10 mL × 2 = 20 mL

However, we need to subtract the initial volume of the solution, which was not provided in the question. Let's assume it was 100 mL:

100 mL + 20 mL = 120 mL

Finally, we need to subtract the volume of HCl that was added before the equivalence point:

120 mL - 114.4 mL = 5.6 mL

Therefore, the answer is B) 5.6 mL.

Question 19

A 0.210-g sample of an acid (molar mass = 192 g/mol)is titrated with [$X] mL of [$Y] M NaOH to a phenolphthalein endpoint.The formula of the acid is:

Accepted Answer

We can use the formula: moles acid = moles base at equivalence point. Moles acid = 0.210 g/molar mass = 0.00109375 mol. Moles base = [$Y] M x [$X] L = [$Y][$X] mol. At equivalence point: 0.00109375 mol acid = [$Y][$X] mol NaOH. We know the molar mass of the acid is 192 g/mol. From the balanced equation: NaOH + HA → NaA + H2O, we know that the mole ratio of NaOH to acid is 1:1, which means that 0.00109375 mol of acid contains 0.00109375 mol of NaOH. Therefore: [$Y][$X] = 0.00109375 mol NaOH / 1 mol acid = 0.00109375 mol / 0.210 g/molar mass = 0.0057 M. Substituting in the values given in the question, [$Y] = 0.0057 M, which is equivalent to a pOH of 2.24. By subtracting this from 14 (pH + pOH = 14), the pH of the solution is 11.76. Phenolphthalein turns pink at a pH of around 8.2-10, indicating that the equivalence point has been reached. The only answer choice that has a potential hydroxide concentration of 0.0057 M is C, with the acid being H₃A.

Question 20

In titrating 100 mL of 0.10 M HCl (aq)with 0.10 M NaOH,the pH does not change a great deal initially.Why is this?

Accepted Answer

The answer of In titrating 100 mL of 0.10 M...

Quiz 15: Acid-Base Equilibria

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 solution containing 10.mmol of and 5.0 mmol of is titrated with 1.7 M HCl.What total volume of HCl must be added to reach the second equivalence point?

A 50.00-mL solution of 0.0350 M hydrocyanic acid (K_a = 6.2 $\times$ 10^-¹⁰) is titrated with a 0.0333 M solution of sodium hydroxide as the titrant.What is the pH of the acid solution after 15.00 mL of titrant have been added? (K_w = 1.00 $\times$ 10^-¹⁴)

A 100.0-mL sample of 0.2 M (CH₃) ₃N (K_b = 5.33 $\times$ 10^-⁵) is titrated with 0.2 M HCl.What is the pH at the equivalence point?

How many of the following will raise the pH of a weak acid HA in aqueous solution? I.addition of water II.making a buffered solution by adding NaA(s) III.addition of NaCl(s) IV.addition of HNO₃ V.titrating with KOH

Calculate the pH at the equivalence point for the titration of 1.0 M ethylamine,C₂H₅NH₂,by 1.0 M perchloric acid,HClO₄.(pK_b for C₂H₅NH₂ = 3.25)

Consider the titration of 100.0 mL of 0.250 M aniline (K_b = 3.8 $\times$ 10^-¹⁰) with 0.500 M HCl.For calculating the volume of HCl required to reach a pH of 8.0,which of the following expressions is correct? (x = volume in mL of HCl required to reach a pH of 8.0)

What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00? The K_a for HOCl is 3.5 $\times$ 10^-⁸.

You dissolve 1.15 grams of an unknown diprotic acid in 200.0 mL of H₂O.This solution is just neutralized by 5.00 mL of a 1.00 M NaOH solution.What is the molar mass of the unknown acid?

Which of the following is a major species present at point IV?

A 2.36-g sample of an acid,H₂X,requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (removing both protons) .The molar mass of the acid is:

A 25.00-mL sample of propanoic acid,CH₃CH₂COOH,of unknown concentration was titrated with 0.111 M KOH.The equivalence point was reached when 44.54 mL of base had been added.What is the concentration of the propanoate ion at the equivalence point?

Which of the following is the net ionic equation for the reaction that occurs during the titration of nitric acid with potassium hydroxide?

Consider the titration of 100.0 mL of 0.250 M aniline (K_b = 3.82 $\times$ 10^-¹⁰) with 0.500 M HCl.Calculate the pH of the solution at the stoichiometric point.

A solution containing 10.mmol of and 5.0 mmol of is titrated with 1.8 M HCl.What volume of HCl must be added to reach the first equivalence point?

A 0.210-g sample of an acid (molar mass = 192 g/mol) is titrated with [$X] mL of [$Y] M NaOH to a phenolphthalein endpoint.The formula of the acid is:

In titrating 100 mL of 0.10 M HCl (aq) with 0.10 M NaOH,the pH does not change a great deal initially.Why is this?

Quiz 15: Acid-Base Equilibria

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 solution containing 10.mmol of and 5.0 mmol of is titrated with 1.7 M HCl.What total volume of HCl must be added to reach the second equivalence point?

A 50.00-mL solution of 0.0350 M hydrocyanic acid (Ka = 6.2 ×\times× 10-10) is titrated with a 0.0333 M solution of sodium hydroxide as the titrant.What is the pH of the acid solution after 15.00 mL of titrant have been added? (Kw = 1.00 ×\times× 10-14)

A 100.0-mL sample of 0.2 M (CH3) 3N (Kb = 5.33 ×\times× 10-5) is titrated with 0.2 M HCl.What is the pH at the equivalence point?

How many of the following will raise the pH of a weak acid HA in aqueous solution? I.addition of water II.making a buffered solution by adding NaA(s) III.addition of NaCl(s) IV.addition of HNO3 V.titrating with KOH

Calculate the pH at the equivalence point for the titration of 1.0 M ethylamine,C2H5NH2,by 1.0 M perchloric acid,HClO4.(pKb for C2H5NH2 = 3.25)

Consider the titration of 100.0 mL of 0.250 M aniline (Kb = 3.8 ×\times× 10-10) with 0.500 M HCl.For calculating the volume of HCl required to reach a pH of 8.0,which of the following expressions is correct? (x = volume in mL of HCl required to reach a pH of 8.0)

What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00? The Ka for HOCl is 3.5 ×\times× 10-8.

You dissolve 1.15 grams of an unknown diprotic acid in 200.0 mL of H2O.This solution is just neutralized by 5.00 mL of a 1.00 M NaOH solution.What is the molar mass of the unknown acid?

Which of the following is a major species present at point IV?

A 2.36-g sample of an acid,H2X,requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (removing both protons) .The molar mass of the acid is:

A 25.00-mL sample of propanoic acid,CH3CH2COOH,of unknown concentration was titrated with 0.111 M KOH.The equivalence point was reached when 44.54 mL of base had been added.What is the concentration of the propanoate ion at the equivalence point?

Which of the following is the net ionic equation for the reaction that occurs during the titration of nitric acid with potassium hydroxide?

Consider the titration of 100.0 mL of 0.250 M aniline (Kb = 3.82 ×\times× 10-10) with 0.500 M HCl.Calculate the pH of the solution at the stoichiometric point.

A solution containing 10.mmol of and 5.0 mmol of is titrated with 1.8 M HCl.What volume of HCl must be added to reach the first equivalence point?

A 0.210-g sample of an acid (molar mass = 192 g/mol) is titrated with [$X] mL of [$Y] M NaOH to a phenolphthalein endpoint.The formula of the acid is:

In titrating 100 mL of 0.10 M HCl (aq) with 0.10 M NaOH,the pH does not change a great deal initially.Why is this?

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 50.00-mL solution of 0.0350 M hydrocyanic acid (K_a = 6.2 $\times$ 10^-¹⁰) is titrated with a 0.0333 M solution of sodium hydroxide as the titrant.What is the pH of the acid solution after 15.00 mL of titrant have been added? (K_w = 1.00 $\times$ 10^-¹⁴)

A 100.0-mL sample of 0.2 M (CH₃) ₃N (K_b = 5.33 $\times$ 10^-⁵) is titrated with 0.2 M HCl.What is the pH at the equivalence point?

How many of the following will raise the pH of a weak acid HA in aqueous solution? I.addition of water II.making a buffered solution by adding NaA(s) III.addition of NaCl(s) IV.addition of HNO₃ V.titrating with KOH

Calculate the pH at the equivalence point for the titration of 1.0 M ethylamine,C₂H₅NH₂,by 1.0 M perchloric acid,HClO₄.(pK_b for C₂H₅NH₂ = 3.25)

Consider the titration of 100.0 mL of 0.250 M aniline (K_b = 3.8 $\times$ 10^-¹⁰) with 0.500 M HCl.For calculating the volume of HCl required to reach a pH of 8.0,which of the following expressions is correct? (x = volume in mL of HCl required to reach a pH of 8.0)

What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00? The K_a for HOCl is 3.5 $\times$ 10^-⁸.

You dissolve 1.15 grams of an unknown diprotic acid in 200.0 mL of H₂O.This solution is just neutralized by 5.00 mL of a 1.00 M NaOH solution.What is the molar mass of the unknown acid?

A 2.36-g sample of an acid,H₂X,requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (removing both protons) .The molar mass of the acid is:

A 25.00-mL sample of propanoic acid,CH₃CH₂COOH,of unknown concentration was titrated with 0.111 M KOH.The equivalence point was reached when 44.54 mL of base had been added.What is the concentration of the propanoate ion at the equivalence point?

Consider the titration of 100.0 mL of 0.250 M aniline (K_b = 3.82 $\times$ 10^-¹⁰) with 0.500 M HCl.Calculate the pH of the solution at the stoichiometric point.