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

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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