Question

A) a solution that is 0.510 M in pyridine (C5H5N) and 0.470 M in pyridinium chloride (C5H5NHCl)

Express your answer using two decimal places.

pH=

B) a solution that is made by combining 55 mL of 5.0×10⁻² M hydrofluoric acid with 125 mL of 0.11 M sodium fluoride

Express your answer using two decimal places.

pH=

C) Calculate the percent ionization of 0.125 M lactic acid in a solution containing 8.5×10⁻³  M sodium lactate.

D) A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 15.0 mL of HNO3.

Express your answer numerically.

pH=

Answer 1

Answer 2

A) To find the pH of the solution containing pyridine and pyridinium chloride, we need to consider the acid-base equilibrium of the pyridine and its conjugate acid, pyridinium ion.

Pyridine (C5H5N) is a weak base, and pyridinium chloride (C5H5NHCl) is its conjugate acid. The equilibrium is as follows:

C5H5N + H2O ⇌ C5H5NH+ + OH-

The concentrations of pyridine (C5H5N) and pyridinium ion (C5H5NH+) are given as 0.510 M and 0.470 M, respectively.

Using the equation for the ion product of water (Kw = [H+][OH-] = 1.0 x 10^-14 at 25°C), we can write:

Kw = [C5H5NH+][OH-]

Now, we can use the fact that [OH-] = [H+] (since they come from the same dissociation of water) to rewrite the equation:

Kw = [C5H5NH+][H+]

Substitute the given concentration of C5H5NH+:

1.0 x 10^-14 = (0.470 M)([H+])

Solve for [H+]:

[H+] = (1.0 x 10^-14) / (0.470 M) ≈ 2.13 x 10^-14 M

Now, calculate the pH:

pH = -log[H+] ≈ -log(2.13 x 10^-14) ≈ 13.67

So, the pH of the solution is approximately 13.67.

B) To find the pH of the solution formed by combining hydrofluoric acid (HF) and sodium fluoride (NaF), we need to consider the reaction of HF with its conjugate base, fluoride ion (F-):

HF + H2O ⇌ H3O+ + F-

Given that the initial volume of hydrofluoric acid (HF) is 55 mL and its concentration is 5.0 x 10^-2 M, we can calculate the number of moles of HF:

moles of HF = (volume of HF in L) x (concentration of HF) = (55 mL / 1000 mL/L) x (5.0 x 10^-2 M) ≈ 2.75 x 10^-3 moles

Next, let's consider the reaction with sodium fluoride (NaF). The initial volume of NaF is 125 mL, and its concentration is 0.11 M. The number of moles of NaF is:

moles of NaF = (volume of NaF in L) x (concentration of NaF) = (125 mL / 1000 mL/L) x (0.11 M) ≈ 1.375 x 10^-2 moles

Since HF and F- react in a 1:1 stoichiometric ratio, the number of moles of HF that react with NaF is limited by the smaller of the two amounts, which is 2.75 x 10^-3 moles. Therefore, 2.75 x 10^-3 moles of HF will react with 2.75 x 10^-3 moles of NaF, leaving 1.375 x 10^-2 - 2.75 x 10^-3 = 1.10 x 10^-2 moles of unreacted NaF.

Now, calculate the total volume of the solution after the reaction:

Total volume = volume of HF + volume of NaF = 55 mL + 125 mL = 180 mL = 0.180 L

Calculate the concentration of H3O+ (hydronium ion) after the reaction:

[H3O+] = (moles of H3O+) / (total volume in L) = (2.75 x 10^-3 moles) / 0.180 L ≈ 1.53 x 10^-2 M

Finally, calculate the pH:

pH = -log[H3O+] ≈ -log(1.53 x 10^-2) ≈ 1.82

So, the pH of the solution is approximately 1.82.

C) To calculate the percent ionization of lactic acid (CH3CH(OH)COOH) in a solution containing sodium lactate (CH3CH(OH)COONa), we need to consider the dissociation of lactic acid into its conjugate base (lactate ion, CH3CH(OH)COO-) and a hydrogen ion (H+):

CH3CH(OH)COOH ⇌ CH3CH(OH)COO- + H+

The percent ionization is defined as the ratio of the concentration of dissociated acid (lactate ion) to the initial concentration of the acid (lactic acid), multiplied by 100:

Percent ionization = ([CH3CH(OH)COO-] / [CH3CH(OH)COOH]) x 100

The initial concentration of lactic acid is given as 0.125 M. The concentration of sodium lactate (CH3CH(OH)COONa) is given as 8.5 x 10^-3 M. Since sodium lactate is a salt formed by the reaction of lactic acid and sodium hydroxide, it completely dissociates into lactate ions and sodium ions:

CH3CH(OH)COONa ⇌ CH3CH(OH)COO- + Na+

Therefore, the concentration of lactate ions ([CH3CH(OH)COO-]) in the solution is 8.5 x 10^-3 M.

Now, calculate the percent ionization:

Percent ionization = (8.5 x 10^-3 M / 0.125 M) x 100 ≈ 6.8%

So, the percent ionization of 0.125 M lactic acid in a solution containing 8.5 x 10^-3 M sodium lactate is approximately 6.8%.

D) To calculate the pH after the addition of 15.0 mL of 0.500 M HNO3 to 75.0 mL of 0.200 M NH3, we need to consider the reaction of ammonia (NH3) with nitric acid (HNO3):

NH3 + HNO3 ⇌ NH4+ + NO3-

First, determine the moles of NH3 and HNO3 in the initial solution:

moles of NH3 = (volume of NH3 in L) x (concentration of NH3) = (75.0 mL / 1000 mL/L) x (0.200 M) = 0.015 moles

moles of HNO3 = (volume of HNO3 in L) x (concentration of HNO3) = (15.0 mL / 1000 mL/L) x (0.500 M) = 0.0075 moles

Since NH3 and HNO3 react in a 1:1 stoichiometric ratio, the amount of NH3 that reacts with HNO3 is limited by the smaller of the two amounts, which is