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
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
