Calculate the ph of the solution resulting from the addition of 85.0 ml of 0.35 m hcl to 30.0 ml of 0.40 m aniline (c6h5nh2). kb (c6h5nh2) = 3.8 x 10-10

pH = 0.81

Explanation:

Step 1: Data given

Volume of a 0.35 M HCl solution = 85.0 mL

Volume of a 0.40 M aniline solution = 30.0 mL

Kb of aniline = 3.8 * 10^-10

Step 2: The balanced equation

C6H5NH2 + HCl → C6H5NH3+ + Cl-

Step 3: Calculate moles

Moles = molarity * volume

Moles HCl = 0.35 M * 0.085 L

Moles HCl = 0.02975 moles

Moles aniline = 0.40 M * 0.030 L

Moles aniline = 0.012 moles

Step 4: Calculates limiting reactant

Aniline is the limiting reactant. It will completely be consumed (0.012 mole)

HCl is in excesS. There will react 0.012 moles. There will remain 0.02975 - 0.012 = 0.01775 moles

Step 5: Calculate molarity HCl

Molarity HCl = moles HCl / total volume

Molarity HCl = 0.01775 moles / 0.115 L

Molarity HCl = 0.154 M

Step 6: Calculate pH

pH = -log[H+]

pH = -log[0.154]

pH = 0.81

explanation: no but i do not usually comment

thats how its supposed to be