Question 1

What is the identity of M in the hydrate M(H₂O)₆ⁿ⁺ that has the 0.10 M solution with the lowest pH?

Accepted Answer

Al^3+ (choice D) forms the most acidic solution because it has the highest charge density among the given options, leading to the strongest attraction to water molecules. This causes the greatest degree of water ionization, releasing more H+ ions and resulting in the lowest pH.

Question 2

Aniline, (C₆H₅NH₂,K_b = 4.3 × 10^-10 at 25°C)is an industrially important amine used in the making of dyes.Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution.

Accepted Answer

Step 1: Write the equation for the dissociation of aniline.
C6H5NH2(aq) + H2O(l) ↔ C6H5NH3+(aq) + OH-(aq)

Step 2: Write the expression for the equilibrium constant (Kb) for the above reaction.
Kb = [C6H5NH3+][OH-]/[C6H5NH2]

Step 3: Calculate the concentration of aniline in the solution.
Moles of aniline = 3.90 g / 93.13 g/mol = 4.18 × 10^-2 mol
Volume of solution = 100 mL = 0.1 L
Concentration of aniline = moles/volume = 4.18 × 10^-2 mol / 0.1 L = 0.418 M

Step 4: Calculate the concentration of hydroxide ions (OH-).
Kb = [C6H5NH3+][OH-]/[C6H5NH2]
Assume that initially, [OH-] = 0 (since aniline is a weak base and will not completely dissociate into ions).
Kb = [C6H5NH3+][OH-]/[C6H5NH2]
4.3 × 10^-10 = x^2 / (0.418 - x)
Assuming x << 0.418, then (0.418 - x) ≈ 0.418.
x = [OH-] = 6.416 × 10^-6 M

Step 5: Calculate the pOH and pH of the solution.
pOH = -log[OH-] = -log(6.416 × 10^-6) = 5.19
pH = 14 - pOH = 8.81

Step 6: Apply the correction factor for the ionization of water.
Since aniline is a weak base, we need to correct for the ionization of water. We can assume that [H+] = [OH-] = 6.416 × 10^-6 M (at 25°C). Therefore, pKa = 14 - pKb = 14 - 4.87 = 9.13
pH = pKa + log([C6H5NH3+]/[C6H5NH2]) = 9.13 + log(0.017) = 9.13 + (-1.77) = 7.36
But remember that we need to apply the correction factor for the ionization of water.
pH = 7.36 + log([H+]/[OH-])
Since [H+] = [OH-] = 6.416 × 10^-6 M,
pH = 7.36 + log(1) = 7.36

Therefore, the pH of the aniline solution is 7.36. The closest answer choice is B) 9.13, but this is not the correct answer.

Question 3

Which one of the following salts,when dissolved in water,produces the solution with a pH closest to 7.00?

Accepted Answer

RbI is a salt formed from a strong acid (HI) and a strong base (RbOH), which typically results in a neutral solution with a pH close to 7.00 when dissolved in water.

Question 4

Identify the Lewis acid that acts as a reactant in the following reaction. Co(H₂O)₆³⁺(aq)+ 6 NH₃(aq)→ Co(NH₃)₆³⁺(aq)+ 6 H₂O(l)

Accepted Answer

The answer of Identify the Lewis acid that acts as...

Question 5

Acetic acid CH₃COOH,has an acid dissociation constant of 1.8 × 10^-5.What is the conjugate base of acetic acid and what is its base dissociation constant?

Accepted Answer

The conjugate base of acetic acid (CH3COOH) is acetate ion (CH3COO-), and the base dissociation constant (Kb) can be calculated using the relation Kw = Ka * Kb, where Kw is the ion product of water (1.0 × 10^-14 at 25°C). Given Ka = 1.8 × 10^-5, Kb = Kw / Ka = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.6 × 10^-10.

Question 6

Calculate the pH of a 0.100 M CH₃NH₃Cl solution.K_b for methylamine,CH₃NH₂,is 3.7 × 10^-4.

Accepted Answer

CH3NH2 is a weak base, and the equation for its reaction with water is:

CH3NH2 + H2O ⇌ CH3NH3+ + OH-

The equilibrium constant expression for this reaction is:

Kb = [CH3NH3+][OH-] / [CH3NH2]

We are given the Kb value as 3.7 × 10-4. We can assume that the concentration of CH3NH3+ at equilibrium is very small compared to the initial concentration of CH3NH2, so we can simplify the expression as:

Kb = [OH-]2 / [CH3NH2]

Solving for [OH-]:

[OH-] = √(Kb × [CH3NH2]) = √(3.7 × 10-4 × 0.100) = 1.21 × 10-3 M

The pH is then calculated as:

pH = 14 - pOH = 14 - (-log[OH-]) = 5.78

Question 7

What is the identity of M in the hydrate M(H₂O)₆ⁿ⁺ that has the 0.10 M solution with the highest pH?

Accepted Answer

In this hydrate, M represents the cation of the salt. The solution with the highest pH would be the one with the cation that behaves as the strongest base, which would be the cation that is the most likely to accept a proton (H+). Among the choices given, Na+ is the most likely to accept a proton, making it the cation with the highest pH solution. Therefore, the answer is B.

Question 8

How many grams of pyridine are there in 100 mL of an aqueous solution that has a pH of 9.00? The K_b for pyridine is 1.9 × 10^-9 and the equation of interest is C₅H₅N(aq)+ H₂O(l)⇌ C₅H₅NH⁺(aq)+ OH^-(aq).

Accepted Answer

First, we need to use the given equation and the Ka value to find the concentration of pyridine in the solution.

Ka = [C5H5NHS+][OH-]/[C5H5N]

At pH 9, [OH-] = 1.0 x 10^-5 M

Let x be the concentration of pyridine in M

1.9 x 10^-9 = (x^2)/(x + 1.0 x 10^-5)

Solving for x, we get x = 1.22 x 10^-4 M

Now, we can use the density of water (1 g/mL) and the molar mass of pyridine (79.1 g/mol) to convert the concentration to grams per mL and then multiply by 100 mL to get the total grams of pyridine:

1.22 x 10^-4 M * 79.1 g/mol * 1 mL/1 g = 0.00965 g/mL

0.00965 g/mL * 100 mL = 0.965 g

However, we need to account for the fact that pyridine is not the only component of the solution - it is dissolved in water and there may be some pyridine hydroxide present. To do this, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([C5H5NHS+]/[C5H5N])

At pH 9, we can assume that [C5H5N] is much larger than [C5H5NHS+], so we can simplify:

9.00 = 8.72 + log([C5H5NHS+]/[C5H5N])

0.28 = log([C5H5NHS+]/[C5H5N])

[C5H5NHS+]/[C5H5N] = 1.93

This means that for every 1.93 moles of pyridine, there is 1 mole of pyridine hydroxide. So we can adjust our concentration of pyridine accordingly:

1.22 x 10^-4 M / 2.93 = 4.15 x 10^-5 M

And now we can calculate the grams of pyridine:

4.15 x 10^-5 M * 79.1 g/mol * 1 mL/1 g = 0.00329 g/mL

0.00329 g/mL * 100 mL = 0.329 g

Therefore, the answer is closest to B) 0.42 g.

Question 9

Equal volumes of 0.10 M NH₃ (K_b = 1.8 × 10^-5)and 0.10 M HCN (K_a = 4.9 × 10^-10)are mixed together.Will the resulting solution be acidic,basic,or neutral?

Accepted Answer

HCN is a weak acid with a pKa value of 9.2, meaning that it dissociates only slightly in water to form H+ and CN- ions. NH4+ is the conjugate acid of a weak base NH3, with a pKa value of 9.3, meaning that it will only slightly dissociate in water to form NH3 and H+ ions. The addition of the NH4+ to the solution will cause a shift in the equilibrium towards the formation of more NH3 and H+ ions. Since NH3 is a weak base, it will not completely neutralize the H+ ions, resulting in a basic solution.

Question 10

What is the pH of a 0.30 M pyridine solution that has a K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is C₅H₅N(aq)+ H₂O(l)⇌ C₅H₅NH⁺(aq)+ OH^-(aq).

Accepted Answer

First, let's write out the equilibrium expression for the dissociation of pyridine:

Kw = [NH+][OH-] = [NH][HPO4-]/[NH2PO4]

The value given, KS1U1B1bS1U1B0 = 1.9 × 10S1U1P1-9S1S1P0, is actually the equilibrium constant for the reaction written in the reverse direction (the association of pyridinium and hydroxide ions to form pyridine and water). To use this value, we need to take the inverse to get the equilibrium constant for the dissociation of pyridine:

Kdiss = 1/KS1U1B1bS1U1B0 = 5.26 × 10S1U1P1S1S1P0

Now, let's set up the ICE table for the dissociation of pyridine:

CS1U1B15S1U1B0HS1U1B15S1U1B0N(aq)+ HS1U1B12S1U1B0O(l)⇌ CS1U1B15S1U1B0HS1U1B15S1U1B0NHS1U1P1+S1S1P0(aq)+ OHS1U1P1-S1S1P0(aq)
I: 0.30 M      0 M              0 M
C: -x          +x              +x
E: 0.30-x      x               x

Next, we can plug the equilibrium concentrations into the expression for Kdiss and solve for x:

Kdiss = [NH+][OH-]/[NH]
5.26 × 10S1U1P1S1S1P0 = xS1S1P0S2/(0.30-x)
x = 1.80 × 10S1U1P1S1S1P0

Finally, we can calculate the pH:

pH = -log[H+]
[H+] = x = 1.80 × 10S1U1P1S1S1P0 M
pH = 9.38

Therefore, the answer is choice C.

Question 11

If you know K_b for ammonia,NH₃,you can calculate the equilibrium constant,K_a,for the following reaction: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ Using the equation:

Accepted Answer

According to the equation provided, the equilibrium constant (K_a) is equal to the ratio of the forward rate constant (K_w) and the reverse rate constant (K_b). Therefore, the correct choice is B, which rearranges the equation accordingly. Choices A, C, and D have incorrect rearrangements of the equation.

Question 12

Calculate the pH of a 0.100 M NaCH₃CO₂ solution.K_a for acetic acid,CH₃CO₂H,is 1.8 × 10^-5.

Accepted Answer

NaCH₃CO₂ is a salt of a weak acid (CH₃CO₂H) and a strong base (NaOH). It hydrolyzes in water to form CH₃CO₂⁻, which reacts with water to form OH⁻, making the solution basic. Using the formula for the hydrolysis of a salt of a weak acid, pH = 14 - 1/2(pKa - log C), where C is the concentration, we find pH ≈ 8.87.

Question 13

Identify the Lewis acid that acts as a reactant in the following reaction. Fe(H₂O)₆³⁺(aq)+ 6 CN^-(aq)→ Fe(CN)₆^3-(aq)+ 6 H₂O(l)

Accepted Answer

The answer of Identify the Lewis acid that acts as...

Question 14

Ammonia NH₃,has a base dissociation constant of 1.8 × 10^-5.What is the conjugate acid of ammonia and what is its acid dissociation constant?

Accepted Answer

The conjugate acid of ammonia (NH₃) is ammonium (NH₄⁺). The acid dissociation constant (Ka) can be found using the relation Kw = Ka × Kb, where Kw is the ion-product constant of water (1.0 × 10⁻¹⁴ at 25°C). Solving for Ka gives 5.6 × 10⁻¹⁰.

Question 15

Determine the ammonia concentration of an aqueous solution that has a pH of 11.50.The equation for the dissociation of NH₃ (K_b = 1.8 × 10^-5)is NH₃(aq)+ H₂O(l)⇌ NH₄⁺(aq)+ OH^-(aq).

Accepted Answer

The dissociation of ammonia in water is given by the following equation:

NH3 + H2O ⇌ NH4+ + OH-

The equilibrium constant for this equation, known as the base dissociation constant (Kb), is given by:

Kb = [NH4+][OH-]/[NH3]

Since pH + pOH = 14 and pOH can be calculated from the given pH, we can solve for [OH-]:

pOH = 14 - pH = 2.50
[OH-] = 10^-pOH = 3.16 × 10^-3 M

Using the equation for Kb and the value of [OH-], we can solve for [NH3]:

Kb = [NH4+][OH-]/[NH3]
1.8 × 10^-5 = x^2 / (0.050 - x)

Solving for x gives:

x = [NH3] = 0.55 M

Therefore, the answer is B).

Question 16

Arrange the following 0.10 M aqueous solutions in order of increasing pH: NaOH,HBr,NaCH₃CO₂,KBr,NH₄Br.

Accepted Answer

HBr is a strong acid, NH4Br is acidic due to NH4+, KBr is neutral, NaCH3CO2 is basic due to CH3CO2-, and NaOH is a strong base.

Question 17

Methylamine CH₃NH₂,has a base dissociation constant of 3.7 × 10^-4.What is the conjugate acid of methylamine and what is its acid dissociation constant?

Accepted Answer

The conjugate acid of methylamine (CH3NH2) is formed by adding a proton (H+), resulting in CH3NH3+. The acid dissociation constant (Ka) for the conjugate acid can be found using the relationship Kw = Ka * Kb, where Kw is the ion product of water (1.0 × 10^-14 at 25°C). Given that the base dissociation constant (Kb) for methylamine is 3.7 × 10^-4, we can calculate Ka as (1.0 × 10^-14) / (3.7 × 10^-4) = 2.7 × 10^-11.

Question 18

If an equal number of moles of the weak acid HCN and the strong base KOH are added to water,is the resulting solution acidic,basic,or neutral?

Accepted Answer

When HCN and KOH are mixed in equal moles, they react in a 1:1 ratio to form KCN and water. KCN is a salt of a strong base (KOH) and a weak acid (HCN), therefore it hydrolyzes in water to produce OH- ions, making the solution basic. Hence, the resulting solution is basic.

Question 19

Dihydrogen phosphate H₂PO₄^-,has an acid dissociation constant of 6.2 × 10^-7.What is the conjugate base of H₂PO₄^- and what is its base dissociation constant?

Accepted Answer

The conjugate base of dihydrogen phosphate (H₂PO₄⁻) is hydrogen phosphate (HPO₄²⁻). The base dissociation constant (Kb) can be calculated using the relation Kw = Ka * Kb, where Kw is the ion product of water (1.0 × 10⁻¹⁴ at 25°C). Given Ka = 6.2 × 10⁻⁷, Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (6.2 × 10⁻⁷) = 1.6 × 10⁻⁸.

Question 20

Calculate the pH of a of 0.100 M KBrO solution.K_a for hypobromous acid,HBrO,is 2.0 × 10^-9.

Accepted Answer

The answer of Calculate the pH of a of 0.100...

Quiz 15: Aqueous Equilibria: Acids and Bases

What is the identity of M in the hydrate M(H₂O) ₆ⁿ⁺ that has the 0.10 M solution with the lowest pH?

Aniline, (C₆H₅NH₂,K_b = 4.3 × 10^-10 at 25°C) is an industrially important amine used in the making of dyes.Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution.

Which one of the following salts,when dissolved in water,produces the solution with a pH closest to 7.00?

Identify the Lewis acid that acts as a reactant in the following reaction. Co(H₂O) ₆³⁺(aq) + 6 NH₃(aq) → Co(NH₃) ₆³⁺(aq) + 6 H₂O(l)

Acetic acid CH₃COOH,has an acid dissociation constant of 1.8 × 10^-5.What is the conjugate base of acetic acid and what is its base dissociation constant?

Calculate the pH of a 0.100 M CH₃NH₃Cl solution.K_b for methylamine,CH₃NH₂,is 3.7 × 10^-4.

What is the identity of M in the hydrate M(H₂O) ₆ⁿ⁺ that has the 0.10 M solution with the highest pH?

How many grams of pyridine are there in 100 mL of an aqueous solution that has a pH of 9.00? The K_b for pyridine is 1.9 × 10^-9 and the equation of interest is C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq) .

Equal volumes of 0.10 M NH₃ (K_b = 1.8 × 10^-5) and 0.10 M HCN (K_a = 4.9 × 10^-10) are mixed together.Will the resulting solution be acidic,basic,or neutral?

What is the pH of a 0.30 M pyridine solution that has a K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq) .

If you know K_b for ammonia,NH₃,you can calculate the equilibrium constant,K_a,for the following reaction: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ Using the equation:

Calculate the pH of a 0.100 M NaCH₃CO₂ solution.K_a for acetic acid,CH₃CO₂H,is 1.8 × 10^-5.

Identify the Lewis acid that acts as a reactant in the following reaction. Fe(H₂O) ₆³⁺(aq) + 6 CN^-(aq) → Fe(CN) ₆^3-(aq) + 6 H₂O(l)

Ammonia NH₃,has a base dissociation constant of 1.8 × 10^-5.What is the conjugate acid of ammonia and what is its acid dissociation constant?

Determine the ammonia concentration of an aqueous solution that has a pH of 11.50.The equation for the dissociation of NH₃ (K_b = 1.8 × 10^-5) is NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) .

Arrange the following 0.10 M aqueous solutions in order of increasing pH: NaOH,HBr,NaCH₃CO₂,KBr,NH₄Br.

Methylamine CH₃NH₂,has a base dissociation constant of 3.7 × 10^-4.What is the conjugate acid of methylamine and what is its acid dissociation constant?

If an equal number of moles of the weak acid HCN and the strong base KOH are added to water,is the resulting solution acidic,basic,or neutral?

Dihydrogen phosphate H₂PO₄^-,has an acid dissociation constant of 6.2 × 10^-7.What is the conjugate base of H₂PO₄^- and what is its base dissociation constant?

Calculate the pH of a of 0.100 M KBrO solution.K_a for hypobromous acid,HBrO,is 2.0 × 10^-9.

Quiz 15: Aqueous Equilibria: Acids and Bases

What is the identity of M in the hydrate M(H2O) 6n+ that has the 0.10 M solution with the lowest pH?

Aniline, (C6H5NH2,Kb = 4.3 × 10-10 at 25°C) is an industrially important amine used in the making of dyes.Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution.

Which one of the following salts,when dissolved in water,produces the solution with a pH closest to 7.00?

Identify the Lewis acid that acts as a reactant in the following reaction. Co(H2O) 63+(aq) + 6 NH3(aq) → Co(NH3) 63+(aq) + 6 H2O(l)

Acetic acid CH3COOH,has an acid dissociation constant of 1.8 × 10-5.What is the conjugate base of acetic acid and what is its base dissociation constant?

Calculate the pH of a 0.100 M CH3NH3Cl solution.Kb for methylamine,CH3NH2,is 3.7 × 10-4.

What is the identity of M in the hydrate M(H2O) 6n+ that has the 0.10 M solution with the highest pH?

How many grams of pyridine are there in 100 mL of an aqueous solution that has a pH of 9.00? The Kb for pyridine is 1.9 × 10-9 and the equation of interest is C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH-(aq) .

Equal volumes of 0.10 M NH3 (Kb = 1.8 × 10-5) and 0.10 M HCN (Ka = 4.9 × 10-10) are mixed together.Will the resulting solution be acidic,basic,or neutral?

What is the pH of a 0.30 M pyridine solution that has a Kb = 1.9 × 10-9? The equation for the dissociation of pyridine is C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH-(aq) .

If you know Kb for ammonia,NH3,you can calculate the equilibrium constant,Ka,for the following reaction: NH4+ + H2O ⇌ NH3 + H3O+ Using the equation:

Calculate the pH of a 0.100 M NaCH3CO2 solution.Ka for acetic acid,CH3CO2H,is 1.8 × 10-5.

Identify the Lewis acid that acts as a reactant in the following reaction. Fe(H2O) 63+(aq) + 6 CN-(aq) → Fe(CN) 63-(aq) + 6 H2O(l)

Ammonia NH3,has a base dissociation constant of 1.8 × 10-5.What is the conjugate acid of ammonia and what is its acid dissociation constant?

Determine the ammonia concentration of an aqueous solution that has a pH of 11.50.The equation for the dissociation of NH3 (Kb = 1.8 × 10-5) is NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) .

Arrange the following 0.10 M aqueous solutions in order of increasing pH: NaOH,HBr,NaCH3CO2,KBr,NH4Br.

Methylamine CH3NH2,has a base dissociation constant of 3.7 × 10-4.What is the conjugate acid of methylamine and what is its acid dissociation constant?

If an equal number of moles of the weak acid HCN and the strong base KOH are added to water,is the resulting solution acidic,basic,or neutral?

Dihydrogen phosphate H2PO4-,has an acid dissociation constant of 6.2 × 10-7.What is the conjugate base of H2PO4- and what is its base dissociation constant?

Calculate the pH of a of 0.100 M KBrO solution.Ka for hypobromous acid,HBrO,is 2.0 × 10-9.

What is the identity of M in the hydrate M(H₂O) ₆ⁿ⁺ that has the 0.10 M solution with the lowest pH?

Aniline, (C₆H₅NH₂,K_b = 4.3 × 10^-10 at 25°C) is an industrially important amine used in the making of dyes.Determine the pH of an aniline solution made by dissolving 3.90 g of aniline in enough water to make 100 mL of solution.

Identify the Lewis acid that acts as a reactant in the following reaction. Co(H₂O) ₆³⁺(aq) + 6 NH₃(aq) → Co(NH₃) ₆³⁺(aq) + 6 H₂O(l)

Acetic acid CH₃COOH,has an acid dissociation constant of 1.8 × 10^-5.What is the conjugate base of acetic acid and what is its base dissociation constant?

Calculate the pH of a 0.100 M CH₃NH₃Cl solution.K_b for methylamine,CH₃NH₂,is 3.7 × 10^-4.

What is the identity of M in the hydrate M(H₂O) ₆ⁿ⁺ that has the 0.10 M solution with the highest pH?

How many grams of pyridine are there in 100 mL of an aqueous solution that has a pH of 9.00? The K_b for pyridine is 1.9 × 10^-9 and the equation of interest is C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq) .

Equal volumes of 0.10 M NH₃ (K_b = 1.8 × 10^-5) and 0.10 M HCN (K_a = 4.9 × 10^-10) are mixed together.Will the resulting solution be acidic,basic,or neutral?

What is the pH of a 0.30 M pyridine solution that has a K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq) .

If you know K_b for ammonia,NH₃,you can calculate the equilibrium constant,K_a,for the following reaction: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ Using the equation:

Calculate the pH of a 0.100 M NaCH₃CO₂ solution.K_a for acetic acid,CH₃CO₂H,is 1.8 × 10^-5.

Identify the Lewis acid that acts as a reactant in the following reaction. Fe(H₂O) ₆³⁺(aq) + 6 CN^-(aq) → Fe(CN) ₆^3-(aq) + 6 H₂O(l)

Ammonia NH₃,has a base dissociation constant of 1.8 × 10^-5.What is the conjugate acid of ammonia and what is its acid dissociation constant?

Determine the ammonia concentration of an aqueous solution that has a pH of 11.50.The equation for the dissociation of NH₃ (K_b = 1.8 × 10^-5) is NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) .

Arrange the following 0.10 M aqueous solutions in order of increasing pH: NaOH,HBr,NaCH₃CO₂,KBr,NH₄Br.

Methylamine CH₃NH₂,has a base dissociation constant of 3.7 × 10^-4.What is the conjugate acid of methylamine and what is its acid dissociation constant?

Dihydrogen phosphate H₂PO₄^-,has an acid dissociation constant of 6.2 × 10^-7.What is the conjugate base of H₂PO₄^- and what is its base dissociation constant?

Calculate the pH of a of 0.100 M KBrO solution.K_a for hypobromous acid,HBrO,is 2.0 × 10^-9.