Question

A volume of 25.0 mL of 0.100 M HCl is titrated against a 0.100 M CH3NH2 solution added to it from a buret.(a) Calculate the pH value of the solution after 10.0 mL of CH3NH2 solution has been added.(b) Calculate the pH value of the solution after 25.0 mL of CH3NH2 solution have been added.(c) Calculate the pH value of the solution after 35.0 mL of CH3NH2 solution have been added.

Answer 1

To calculate the pH value after adding a certain volume of CH3NH2 solution to the 25.0 mL of 0.100 M HCl solution, we need to consider the reaction between the HCl and CH3NH2. After adding 10.0 mL, 25.0 mL, and 35.0 mL of CH3NH2 solution, we can calculate the pH value using the Henderson-Hasselbalch equation and considering the hydrolysis of the CH3NH3Cl salt.

Step-by-step explanation:

To calculate the pH value after adding a certain volume of CH3NH2 solution to the 25.0 mL of 0.100 M HCl solution, we need to consider the reaction between the HCl and CH3NH2. The reaction between a strong acid (HCl) and a weak base (CH3NH2) forms the salt CH3NH3Cl. Since CH3NH3Cl is a weak acid, it will undergo hydrolysis in water to produce H3O+ ions and the conjugate base CH3NH2. In this case, we are adding CH3NH2 solution to the HCl solution, so the pH of the solution will be less acidic than the original HCl solution.

a) After adding 10.0 mL of CH3NH2 solution, we have a total volume of 35.0 mL. To calculate the pH, we can use the equation pH = -log[H3O+]. Since we are dealing with a weak acid, we also need to consider the hydrolysis of the CH3NH3Cl salt. The pH value can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. The pKa value for CH3NH3Cl is given as 4.75, and the concentration of CH3NH3+ can be calculated using the initial concentration and the volume of CH3NH2 solution added. From there, we can calculate the concentration of CH3NH2 and use the Henderson-Hasselbalch equation to find the pH value.

b) After adding 25.0 mL of CH3NH2 solution, we have a total volume of 50.0 mL. Using the same approach as in part a), we can calculate the pH value considering the hydrolysis of the CH3NH3Cl salt.

c) After adding 35.0 mL of CH3NH2 solution, we have a total volume of 60.0 mL. Using the same approach as in parts a) and b), we can calculate the pH value considering the hydrolysis of the CH3NH3Cl salt.

Answer 2

a) pH = 1.37

b) pH = 5.76

c) pH = 10.24

Step-by-step explanation:

Step 1: Data given

Volume of a 0.100 M HCl = 25.0 mL = 0.025 L

Molarity of a CH3NH2 solution = 0.100 M

Step 2: The balanced equation

2CH3NH2 + 2HCl → 2CH3NH3 + Cl2

Calculate the pH value of the solution after 10.0 mL of CH3NH2 solution has been added.

Step 3: Calculate moles

Moles HCl = molarity * volume

Moles HCl = 0.100 M * 0.025 L

Moles HCl = 0.0025 moles

Moles CH3NH2 = 0.100 M * 0.010 L

Moles CH3NH2 = 0.001 moles

Step 4: Calculate the limiting reactant

For 2 moles CH3NH2 we need 2 moles HCl to produce 2 moles CH3NH3 and 1 mol Cl2

CH3NH2 is the limiting reactant. There will react 0.001 moles. HCl is in excess. There will react 0.001 moles. There will remain 0.0025 - 0.001 = 0.0015 moles HCl

Step 5: Calculate molarity HCl

Molarity = moles / volume

Molarity = 0.0015 moles / 0.035 L

Molarity = 0.043 M

Step 6: Calculate pH

pH HCl = -log[H+]

pH HCl = -log( 0.043)

pH HCl = 1.37

Calculate the pH value of the solution after 25.0 mL of CH3NH2 solution have been added.

Step 3: Calculate moles

Moles HCl = molarity * volume

Moles HCl = 0.100 M * 0.025 L

Moles HCl = 0.0025 moles

Moles CH3NH2 = 0.100 M * 0.025 L

Moles CH3NH2 = 0.0025 moles

Step 4: Calculate the limiting reactant

For 2 moles CH3NH2 we need 2 moles HCl to produce 2 moles CH3NH3 and 1 mol Cl2

Both reactants will completely be consumed.

Ka = Kw/Kb = 1.0 * 10^-14 / 4.2 * 10^-4 =2.38 * 10^-11 = x² / 0.125-x

x = [H+]= 1.73 * 10^-6 M

pH = -log[H+]

pH = -log( 1.73 * 10^-6)

pH = 5.76

Calculate the pH value of the solution after 35.0 mL of CH3NH2 solution have been added.

Step 3: Calculate moles

Moles HCl = molarity * volume

Moles HCl = 0.100 M * 0.025 L

Moles HCl = 0.0025 moles

Moles CH3NH2 = 0.100 M * 0.035 L

Moles CH3NH2 = 0.0035 moles

Step 4: Calculate the limiting reactant

For 2 moles CH3NH2 we need 2 moles HCl to produce 2 moles CH3NH3 and 1 mol Cl2

There will remain 0.0010 moles of CH3NH2

There will be 0.0025 mole of CH3NH3 be produced

Step 5: Calculate the molarity

[CH3NH2] = 0.0010 moles / 0.060 L

[CH3NH2]= 0.0167 M

[CH3NH3] = 0.0025 moles / 0.060 L

[CH3NH3] = 0.0417 M

Step 5: Calculate pOH

pOH = pKb + log ([B+]/[BOH]

⇒[B+] = the concentration of CH3NH3 = 0.0417 M

⇒[BOH] =the concentration of CH3NH2= 0.0167 M

⇒pKb = 3.36

pOH = 3.36 + log (0.0417 / 0.0167)

pOH = 3.36 + 0.40

pOH = 3.76

pH = 14 - 3.76

pH = 10.24