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.

a) pH = 1.37

b) pH = 5.76

c) pH = 10.24

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

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.

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