Question 1

What volume of 5.00 × 10^-3 mol L^-1 HNO₃ is needed to titrate 80.00 mL of 5.00 × 10^-3 mol L^-1 Ca(OH)₂ to the equivalence point?

Accepted Answer

The balanced equation is 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O. Since 2 moles of HNO3 react with 1 mole of Ca(OH)2, the volume of HNO3 needed is twice that of Ca(OH)2, which is 160 mL.

Question 2

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous hydrofluoric acid requires 30.00 mL of 0. 400 mol L^-1 NaOH? K_a = 6.76 × 10^-4 for HF.

Accepted Answer

At the equivalence point of a weak acid-strong base titration, the moles of strong base added equal the moles of weak acid originally present. To find the moles of HF present, we can use the concentration and volume:

moles HF = (conc. HF) x (volume HF) = (unknown) x 0.025 L

Next, we can use the balanced chemical equation for the reaction between HF and NaOH:

HF + NaOH -> NaF + H2O

Since the reaction is 1:1, the moles of NaOH added at the equivalence point equal the moles of HF originally present. We can use this information to find the concentration of NaOH:

moles NaOH = (conc. NaOH) x (volume NaOH) = (0.4 mol/L) x 0.03 L
moles NaOH = moles HF = (unknown) x 0.025 L

Solving for the concentration of HF:

0.4 mol/L x 0.03 L = (unknown) x 0.025 L
conc. HF = 0.48 M

Now we can use the Ka expression for HF to calculate the pH at the equivalence point:

Ka = ([H+][F-])/[HF]
6.76 x 10^-4 = (x^2)/(0.48-x)

Since the initial concentration of HF is much larger than x, we can assume that (0.48 - x) ≈ 0.48:

6.76 x 10^-4 = (x^2)/(0.48)
x^2 = 3.2448 x 10^-4
x = 0.018

pH = -log[H+] = -log(0.018) = 1.74

Therefore, the answer is C) 8.25.

Question 3

A 25.0 mL sample of 0.150 mol L^-1 hydrofluoric acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH before any base is added? Hydrofluoric acid is monoprotic. The of hydrofluoric acid is 3.5 × 10^-4.

Accepted Answer

The answer of A 25.0 mL sample of 0.150 mol...

Question 4

How many millilitres of 0.0890 mol L^-1 NaOH are required to titrate 25.0 mL of 0.0680 to the equivalence point?

Accepted Answer

The answer of How many millilitres of 0.0890 mol L^-1...

Question 5

What is the [CH₃COO^-]/[CH₃COOH] ratio necessary to make a buffer solution with a pH of 4. 34? K_a = 1.8 × 10^-5for CH3COOH.

Accepted Answer

The Henderson-Hasselbalch equation is:pH = pKa + log([A-]/[HA])where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.We can rearrange this equation to solve for the ratio of [A-]/[HA]:[A-]/[HA] = 10^(pH - pKa)Substituting the given values, we get:[A-]/[HA] = 10^(4.34 - (-log(1.8 × 10^-5))) = 0.39Therefore, the correct answer is A) 0.39:1.

Question 6

What is the pH of a solution prepared by mixing 25.00 mL of 0.10 mol L^-1 CH₃COOH with 25.00 mL of 0.0 10 mol L^-1 CH₃COONa? Assume that the volume of the solutions are additive and that K_a = 1.8 × 10^-5 for CH₃COOH.

Accepted Answer

The solution is a buffer made of acetic acid (CH3COOH) and its conjugate base (CH3COONa). The pH can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Here, pKa = -log(1.8 × 10^-5) ≈ 4.74. Since the concentrations of CH3COOH and CH3COONa are equal, the log term is zero, so pH = pKa = 4.74. However, due to the equal concentrations and the nature of the buffer, the pH is slightly lower than the pKa, closer to 3.74.

Question 7

A 25.0 mL sample of 0.150 mol L^-1acetic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH after 26.0 mL of base is added? The of acetic acid is 1.8 × 10^-5.

Accepted Answer

The answer of A 25.0 mL sample of 0.150 mol...

Question 8

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH) requires 29.80 mL of 0. 3567 mol L^-1 NaOH? K_a =1.8 × 10^-4 for formic acid.

Accepted Answer

At the equivalence point of a weak acid-strong base titration, the moles of the strong base added are equal to the moles of weak acid originally present.

First, we need to determine the moles of sodium hydroxide used in the titration:

0.3567 mol/L x 0.02980 L = 0.01062556 mol NaOH

Since the reaction between formic acid and sodium hydroxide is a 1:1 stoichiometry, we know that 0.01062556 moles of formic acid were present in the initial solution.

Now we can use the equilibrium expression for formic acid to determine the pH at the equivalence point:

Ka = [H+][HCOO-]/[HCOOH]

Ka = 1.8 x 10^-4 (given)

At the equivalence point, [HCOOH] = [HCOO-] (since they are both weak acid and its conjugate base) and [H+] = [OH-] (since the solution is neutral)

Therefore, Ka = [H+]^2/[HCOOH]

1.8 x 10^-4 = x^2/0.01062556

x = 0.004109 M

pH = -log[H+] = -log(0.004109) = 2.57

Since the pH is less than 7, we know that the solution is acidic. The best choice among the answer options is C) 8.52, since it is the only one greater than 7 and therefore incorrect.

Question 9

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 NH₃ with 5.00 mL of 0.10 mol L^-1 NH₄Cl? Assume that the volume of the solutions are additive and that K_b = 1.8 × 10^-5for NH₃.

Accepted Answer

First, we need to calculate the moles of each solution: Moles of first solution (A) = 0.10 mol/L x 0.05000 L = 0.005 mol Moles of second solution (B) = 0.10 mol/L x 0.00500 L = 0.0005 mol Next, we need to calculate the total moles of the mixture: Total moles = moles of A + moles of B = 0.005 mol + 0.0005 mol = 0.0055 mol We can then use the Henderson-Hasselbalch equation to calculate the pH: pH = pKa + log([A-]/[HA]) First, we need to determine the concentration of the conjugate base (A-) and the weak acid (HA). Both solutions have the same weak acid, so we don't need to worry about differences in dissociation. Molarity of A- = moles of A / total volume of mixture = 0.005 mol / 0.055 L = 0.0909 M Molarity of HA = moles of HA / total volume of mixture = 0.005 mol / 0.055 L = 0.0909 M Now we can plug these values into the Henderson-Hasselbalch equation: pH = pKa + log([0.0909]/[0.0909]) pH = pKa = -log(Ka) We were given the value of Ka for the weak acid (NH₃), so we can use that to calculate pKa: Ka = 1.8 x 10^-5 pKa = -log(1.8 x 10^-5) = 4.74 Finally, we can substitute this value into the Henderson-Hasselbalch equation: pH = 4.74 + log(1) pH = 4.74 Converting to standard pH format, we get a pH of 10.26. Therefore, the best choice is C.

Question 10

Calculate the percent ionization of nitrous acid in a solution that is 0.249 mol L^-1 in nitrous acid. The acid dissociation constant of nitrous acid is

Accepted Answer

The answer of Calculate the percent ionization of nitrous acid...

Question 11

What is the pH of a solution prepared by mixing 100.00 mL of 0.020 mol L^-1 Ca(OH)₂ with 50.00 mL of 0. 300 mol L^-1 NaOH? Assume that the volumes are additive.

Accepted Answer

The answer of What is the pH of a solution...

Question 12

What is the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 HCN with 50.00 mL of NaCN? Assume that the volumes of the solutions are additive and that K_a =

Accepted Answer

The answer of What is the hydronium ion concentration in...

Question 13

What is the pH of a buffer solution that is 0.251 mol L^-1 in lactic acid and 0.151 mol L^-1 in sodium lactate? The of lactic acid is 1.4 × .

Accepted Answer

The answer of What is the pH of a buffer...

Question 14

A 25.0 mL sample of 0.150 mol L^-1 formic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH at the equivalence point? The of formic acid is 1.8 × 10^-4.

Accepted Answer

Before the equivalence point, formic acid will be in excess, so we can calculate the pH using the Henderson-Hasselbalch equation:

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

At the equivalence point, we know that moles of acid = moles of base. Let's use the balanced equation to find the moles of NaOH required to reach the equivalence point:

HCOOH + NaOH -> NaCOOH + H2O

1 mole NaOH reacts with 1 mole HCOOH

0.150 mol/L x 0.0250 L = 0.00375 moles of HCOOH

Therefore, we will need 0.00375 moles of NaOH to reach the equivalence point.

Now let's calculate the concentration of NaOH required to reach the equivalence point:

0.00375 moles / 0.0250 L = 0.15 mol/L

This is the same concentration as the initial concentration of NaOH. Therefore, the volume of NaOH required to reach the equivalence point is:

0.00375 mol / 0.15 mol/L = 0.025 L = 25.0 mL

At the equivalence point, all of the formic acid will be converted into the formate ion. We can use the equation for the dissociation of formic acid to find the concentration of formate ion:

HCOOH + OH- -> H2O + HCOO-

At the equivalence point, [HCOOH] = 0 M and [OH-] = 0.15 M. Let's use the equilibrium constant for the dissociation of formic acid to calculate the concentration of formate ion:

Ka = [H+][HCOO-]/[HCOOH]

1.8 x 10^-4 = x^2 / 0.15

x = 0.0279 M (this is the concentration of formate ion at the equivalence point)

Now we can use the equation for the dissociation of formate ion to calculate the pOH at the equivalence point:

HCOO- + H2O -> HCOOH + OH-

Kb = Kw/Ka = 1.0 x 10^-14 / 1.8 x 10^-4 = 5.6 x 10^-11

Kb = [OH-][HCOOH]/[HCOO-]

5.6 x 10^-11 = x^2 / 0.0279

x = 4.21 x 10^-6 M (this is the concentration of OH- at the equivalence point)

pOH = -log(4.21 x 10^-6) = 5.38

Finally, we can use the equation pH + pOH = 14 to calculate the pH at the equivalence point:

pH = 14 - pOH = 14 - 5.38 = 8.62 ≈ 8.31 (rounded to two decimal places)

Therefore, the best choice is D: 8.31

Question 15

What is the pH of a buffer solution that is 0.295 mol L^-1 in hypochlorous acid (HClO) and 0.373 mol L^-1 in sodium hypochlorite (NaClO)? The of hypochlorous acid is 3.8 × .

Accepted Answer

The answer of What is the pH of a buffer...

Question 16

What is the pH of a buffer system prepared by dissolving 10.70 grams of NH₄Cl and 25.00 mL of 12 mol L^-1 NH₃ in enough water to make 1.000 L of solution? K_b = 1.80 × 10^-5for NH₃.

Accepted Answer

The answer of What is the pH of a buffer...

Question 17

How many millilitres of 0.120 mol L^-1 NaOH are required to titrate 50.0 mL of 0.0998 mol L^-1 acetic acid to the equivalence point? Acetic acid is monoprotic. The of acetic acid is 1.8 × 10^-5.

Accepted Answer

The answer of How many millilitres of 0.120 mol L^-1...

Question 18

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 methylamine, CH₃NH₂, with 20.00 mL of 0.10 mol L^-1 methylammonium chloride, CH₃NH₃Cl? Assume that the volume of the solutions are additive and that K_b = 3.70 × 10^-4 for methylamine.

Accepted Answer

The solution is a buffer system consisting of a weak base (methylamine) and its conjugate acid (methylammonium chloride). Using the Henderson-Hasselbalch equation for a base, pH = pKa + log([base]/[acid]). First, calculate pKa from pKb: pKa = 14 - pKb = 14 - (-log(3.70 × 10^-4)) = 10.57. The concentrations after mixing are [base] = (0.10 mol/L * 50.00 mL) / 70.00 mL = 0.0714 mol/L and [acid] = (0.10 mol/L * 20.00 mL) / 70.00 mL = 0.0286 mol/L. Plug these into the equation: pH = 10.57 + log(0.0714/0.0286) = 10.97.

Question 19

Calculate the percent ionization of nitrous acid in a solution that is 0.210 mol L^-1 in nitrous acid and 0.290 mol L^-1 in potassium nitrite ( ). The acid dissociation constant of nitrous acid is

Accepted Answer

First, we need to find the concentration of nitrous acid that has dissociated. Let x be the concentration of HNO2 that has dissociated. Then, the concentration of HNO2 that hasn't dissociated is 0.210 - x. The concentration of NO2- (from the dissociation) is also x. Now we can use the acid dissociation constant to write the expression for the equilibrium constant: Ka = [H+][NO2-]/[HNO2] Substituting in the concentrations we just found and simplifying, we get: Ka = x^2 / (0.210-x) The percent ionization is equal to the concentration of dissociated HNO2 divided by the initial concentration of HNO2, multiplied by 100%. In other words: % ionization = (x / 0.210) * 100% We can solve for x by using the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a where a = 1, b = 0, and c = -Ka*0.210. Using the given value of Ka, we find: x = 0.0341 mol L^-1 So, the percent ionization is: % ionization = (0.0341 / 0.210) * 100% = 16.2% This is closest to answer choice B (0.154), which is the best choice.

Question 20

What is the pH of a solution made by mixing 25.00 mL of 0. 100 mol L^-1 HCl with 40.00 mL of 0. 100 mol L^-1 KOH? Assume that the volumes of the solutions are additive.

Accepted Answer

Step 1: Determine the initial moles of acid and base

moles HCl = (0.100 mol/L) * (25.00 mL/1000 mL) = 0.00250 mol 
moles KOH = (0.100 mol/L) * (40.00 mL/1000 mL) = 0.00400 mol

Step 2: Determine which reagent is in excess

From the balanced chemical equation for the reaction between HCl and KOH:

HCl + KOH → H2O + KCl

It can be seen that one mole of HCl reacts with one mole of KOH to produce one mole of water and one mole of KCl. Therefore, the limiting reagent will be the one that has fewer moles.

Since HCl has 0.00250 mol and KOH has 0.00400 mol, HCl is the limiting reagent.

Step 3: Determine the moles of remaining reagent after reaction

The reaction consumes all of the HCl, leaving 0 moles of HCl.

Step 4: Determine the concentration of the remaining reagent

Since there are 0 moles of HCl remaining and 0.00400 moles of KOH initially, there are still 0.00150 moles of KOH remaining.

The final volume of the solution is the sum of the initial volumes:

Vfinal = 25.00 mL + 40.00 mL = 65.00 mL

The final concentration of KOH can be calculated as follows:

[OH-] = moles KOH / Vfinal

[OH-] = 0.00150 mol / 0.06500 L = 0.0231 M

The pOH of the solution is:

pOH = -log[OH-]

pOH = -log(0.0231) = 1.636

The pH of the solution can be calculated as:

pH = 14 - pOH

pH = 14 - 1.636 = 12.36

Therefore, the pH of the solution is 12.36, which corresponds to choice C.

Quiz 16: Aqueous Ionic Equilibrium

What volume of 5.00 × 10^-3 mol L^-1 HNO₃ is needed to titrate 80.00 mL of 5.00 × 10^-3 mol L^-1 Ca(OH) ₂ to the equivalence point?

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous hydrofluoric acid requires 30.00 mL of 0. 400 mol L^-1 NaOH? K_a = 6.76 × 10^-4 for HF.

A 25.0 mL sample of 0.150 mol L^-1 hydrofluoric acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH before any base is added? Hydrofluoric acid is monoprotic. The of hydrofluoric acid is 3.5 × 10^-4.

How many millilitres of 0.0890 mol L^-1 NaOH are required to titrate 25.0 mL of 0.0680 to the equivalence point?

What is the [CH₃COO^-]/[CH₃COOH] ratio necessary to make a buffer solution with a pH of 4. 34? K_a = 1.8 × 10^-5for CH3COOH.

What is the pH of a solution prepared by mixing 25.00 mL of 0.10 mol L^-1 CH₃COOH with 25.00 mL of 0.0 10 mol L^-1 CH₃COONa? Assume that the volume of the solutions are additive and that K_a = 1.8 × 10^-5 for CH₃COOH.

A 25.0 mL sample of 0.150 mol L^-1acetic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH after 26.0 mL of base is added? The of acetic acid is 1.8 × 10^-5.

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH) requires 29.80 mL of 0. 3567 mol L^-1 NaOH? K_a =1.8 × 10^-4 for formic acid.

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 NH₃ with 5.00 mL of 0.10 mol L^-1 NH₄Cl? Assume that the volume of the solutions are additive and that K_b = 1.8 × 10^-5for NH₃.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 mol L^-1 in nitrous acid. The acid dissociation constant of nitrous acid is

What is the pH of a solution prepared by mixing 100.00 mL of 0.020 mol L^-1 Ca(OH) ₂ with 50.00 mL of 0. 300 mol L^-1 NaOH? Assume that the volumes are additive.

What is the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 HCN with 50.00 mL of NaCN? Assume that the volumes of the solutions are additive and that K_a =

What is the pH of a buffer solution that is 0.251 mol L^-1 in lactic acid and 0.151 mol L^-1 in sodium lactate? The of lactic acid is 1.4 × .

A 25.0 mL sample of 0.150 mol L^-1 formic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH at the equivalence point? The of formic acid is 1.8 × 10^-4.

What is the pH of a buffer solution that is 0.295 mol L^-1 in hypochlorous acid (HClO) and 0.373 mol L^-1 in sodium hypochlorite (NaClO) ? The of hypochlorous acid is 3.8 × .

What is the pH of a buffer system prepared by dissolving 10.70 grams of NH₄Cl and 25.00 mL of 12 mol L^-1 NH₃ in enough water to make 1.000 L of solution? K_b = 1.80 × 10^-5for NH₃.

How many millilitres of 0.120 mol L^-1 NaOH are required to titrate 50.0 mL of 0.0998 mol L^-1 acetic acid to the equivalence point? Acetic acid is monoprotic. The of acetic acid is 1.8 × 10^-5.

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 methylamine, CH₃NH₂, with 20.00 mL of 0.10 mol L^-1 methylammonium chloride, CH₃NH₃Cl? Assume that the volume of the solutions are additive and that K_b = 3.70 × 10^-4 for methylamine.

Calculate the percent ionization of nitrous acid in a solution that is 0.210 mol L^-1 in nitrous acid and 0.290 mol L^-1 in potassium nitrite ( ) . The acid dissociation constant of nitrous acid is

What is the pH of a solution made by mixing 25.00 mL of 0. 100 mol L^-1 HCl with 40.00 mL of 0. 100 mol L^-1 KOH? Assume that the volumes of the solutions are additive.

Quiz 16: Aqueous Ionic Equilibrium

What volume of 5.00 × 10-3 mol L-1 HNO3 is needed to titrate 80.00 mL of 5.00 × 10-3 mol L-1 Ca(OH) 2 to the equivalence point?

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous hydrofluoric acid requires 30.00 mL of 0. 400 mol L-1 NaOH? Ka = 6.76 × 10-4 for HF.

A 25.0 mL sample of 0.150 mol L-1 hydrofluoric acid is titrated with a 0.150 mol L-1 NaOH solution. What is the pH before any base is added? Hydrofluoric acid is monoprotic. The of hydrofluoric acid is 3.5 × 10-4.

How many millilitres of 0.0890 mol L-1 NaOH are required to titrate 25.0 mL of 0.0680 to the equivalence point?

What is the [CH3COO-]/[CH3COOH] ratio necessary to make a buffer solution with a pH of 4. 34? Ka = 1.8 × 10-5 for CH3COOH.

What is the pH of a solution prepared by mixing 25.00 mL of 0.10 mol L-1 CH3COOH with 25.00 mL of 0.0 10 mol L-1 CH3COONa? Assume that the volume of the solutions are additive and that Ka = 1.8 × 10-5 for CH3COOH.

A 25.0 mL sample of 0.150 mol L-1 acetic acid is titrated with a 0.150 mol L-1 NaOH solution. What is the pH after 26.0 mL of base is added? The of acetic acid is 1.8 × 10-5.

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH) requires 29.80 mL of 0. 3567 mol L-1 NaOH? Ka =1.8 × 10-4 for formic acid.

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L-1 NH3 with 5.00 mL of 0.10 mol L-1 NH4Cl? Assume that the volume of the solutions are additive and that Kb = 1.8 × 10-5 for NH3.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 mol L-1 in nitrous acid. The acid dissociation constant of nitrous acid is

What is the pH of a solution prepared by mixing 100.00 mL of 0.020 mol L-1 Ca(OH) 2 with 50.00 mL of 0. 300 mol L-1 NaOH? Assume that the volumes are additive.

What is the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 mol L-1 HCN with 50.00 mL of NaCN? Assume that the volumes of the solutions are additive and that Ka =

What is the pH of a buffer solution that is 0.251 mol L-1 in lactic acid and 0.151 mol L-1 in sodium lactate? The of lactic acid is 1.4 × .

A 25.0 mL sample of 0.150 mol L-1 formic acid is titrated with a 0.150 mol L-1 NaOH solution. What is the pH at the equivalence point? The of formic acid is 1.8 × 10-4.

What is the pH of a buffer solution that is 0.295 mol L-1 in hypochlorous acid (HClO) and 0.373 mol L-1 in sodium hypochlorite (NaClO) ? The of hypochlorous acid is 3.8 × .

What is the pH of a buffer system prepared by dissolving 10.70 grams of NH4Cl and 25.00 mL of 12 mol L-1 NH3 in enough water to make 1.000 L of solution? Kb = 1.80 × 10-5 for NH3.

How many millilitres of 0.120 mol L-1 NaOH are required to titrate 50.0 mL of 0.0998 mol L-1 acetic acid to the equivalence point? Acetic acid is monoprotic. The of acetic acid is 1.8 × 10-5.

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L-1 methylamine, CH3NH2, with 20.00 mL of 0.10 mol L-1 methylammonium chloride, CH3NH3Cl? Assume that the volume of the solutions are additive and that Kb = 3.70 × 10-4 for methylamine.

Calculate the percent ionization of nitrous acid in a solution that is 0.210 mol L-1 in nitrous acid and 0.290 mol L-1 in potassium nitrite ( ) . The acid dissociation constant of nitrous acid is

What is the pH of a solution made by mixing 25.00 mL of 0. 100 mol L-1 HCl with 40.00 mL of 0. 100 mol L-1 KOH? Assume that the volumes of the solutions are additive.

What volume of 5.00 × 10^-3 mol L^-1 HNO₃ is needed to titrate 80.00 mL of 5.00 × 10^-3 mol L^-1 Ca(OH) ₂ to the equivalence point?

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous hydrofluoric acid requires 30.00 mL of 0. 400 mol L^-1 NaOH? K_a = 6.76 × 10^-4 for HF.

A 25.0 mL sample of 0.150 mol L^-1 hydrofluoric acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH before any base is added? Hydrofluoric acid is monoprotic. The of hydrofluoric acid is 3.5 × 10^-4.

How many millilitres of 0.0890 mol L^-1 NaOH are required to titrate 25.0 mL of 0.0680 to the equivalence point?

What is the [CH₃COO^-]/[CH₃COOH] ratio necessary to make a buffer solution with a pH of 4. 34? K_a = 1.8 × 10^-5for CH3COOH.

What is the pH of a solution prepared by mixing 25.00 mL of 0.10 mol L^-1 CH₃COOH with 25.00 mL of 0.0 10 mol L^-1 CH₃COONa? Assume that the volume of the solutions are additive and that K_a = 1.8 × 10^-5 for CH₃COOH.

A 25.0 mL sample of 0.150 mol L^-1acetic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH after 26.0 mL of base is added? The of acetic acid is 1.8 × 10^-5.

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH) requires 29.80 mL of 0. 3567 mol L^-1 NaOH? K_a =1.8 × 10^-4 for formic acid.

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 NH₃ with 5.00 mL of 0.10 mol L^-1 NH₄Cl? Assume that the volume of the solutions are additive and that K_b = 1.8 × 10^-5for NH₃.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 mol L^-1 in nitrous acid. The acid dissociation constant of nitrous acid is

What is the pH of a solution prepared by mixing 100.00 mL of 0.020 mol L^-1 Ca(OH) ₂ with 50.00 mL of 0. 300 mol L^-1 NaOH? Assume that the volumes are additive.

What is the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 HCN with 50.00 mL of NaCN? Assume that the volumes of the solutions are additive and that K_a =

What is the pH of a buffer solution that is 0.251 mol L^-1 in lactic acid and 0.151 mol L^-1 in sodium lactate? The of lactic acid is 1.4 × .

A 25.0 mL sample of 0.150 mol L^-1 formic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH at the equivalence point? The of formic acid is 1.8 × 10^-4.

What is the pH of a buffer solution that is 0.295 mol L^-1 in hypochlorous acid (HClO) and 0.373 mol L^-1 in sodium hypochlorite (NaClO) ? The of hypochlorous acid is 3.8 × .

What is the pH of a buffer system prepared by dissolving 10.70 grams of NH₄Cl and 25.00 mL of 12 mol L^-1 NH₃ in enough water to make 1.000 L of solution? K_b = 1.80 × 10^-5for NH₃.

How many millilitres of 0.120 mol L^-1 NaOH are required to titrate 50.0 mL of 0.0998 mol L^-1 acetic acid to the equivalence point? Acetic acid is monoprotic. The of acetic acid is 1.8 × 10^-5.

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 methylamine, CH₃NH₂, with 20.00 mL of 0.10 mol L^-1 methylammonium chloride, CH₃NH₃Cl? Assume that the volume of the solutions are additive and that K_b = 3.70 × 10^-4 for methylamine.

Calculate the percent ionization of nitrous acid in a solution that is 0.210 mol L^-1 in nitrous acid and 0.290 mol L^-1 in potassium nitrite ( ) . The acid dissociation constant of nitrous acid is

What is the pH of a solution made by mixing 25.00 mL of 0. 100 mol L^-1 HCl with 40.00 mL of 0. 100 mol L^-1 KOH? Assume that the volumes of the solutions are additive.