Question

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

Answer 1

Answer:

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

Answer 2

Answer:

pH = 0.81

Explanation:

HCl reacts with aniline, thus:

C₆H₅NH₂ + HCl → C₆H₅NH₃⁺ + Cl⁻

Moles of HCl are:

0.085L × (0.35mol HCl / L) = 0.02975mol HCl

Moles of aniline are:

0.030L × (0.40mol HCl / L) = 0.012mol aniline

Thus, after reaction, will remain:

0.02975mol - 0.012mol = 0.01775mol HCl

Moles of HCl in solution are equal to moles of H⁺, thus, moles of H⁺ are: 0.01775mol H⁺

As total volume is 85.0mL + 30.0mL = 115.0mL ≡ 0.115L

0.01775mol / 0.115L = 0.1543M

pH of solution = -log[H⁺]

pH = -log 0.1543M

pH = 0.81