Answer :
Answer:
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