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

water   molecules stay close to each other through cohesion, the collective action of hydrogen bonds between water molecules. cohesion plants draw water upwards from the root through the stem's transport tubes (xylem) to the leaf. this transport occurs even in large trees and against gravity. transpiration, evaporation of water from plant leaves, creates a tension on water molecules being pulled up from the stem and roots.

water also has high (the attraction of one substance to another) properties because of its polar nature. adhesion of water molecules to plant cell walls counteract gravity while water is transported from roots to leaves.

water exhibits a very high surface tension when compared to other liquids. surface tension is a property of a liquid that holds the surface together and allows it to resist an external force. by capillary action, the water forms concave menisci inside the pores. the high surface tension of water pulls the concavity outwards, generating enough force to lift water from the roots to the leaves of plants.

thus, cohesion, along with adhesion and surface tension creates a capillary action that keeps water molecules interacting and moving through the plants out to the leaf cells.

therefore, option d) is the correct answer.

explanation: