Answer:
8.03.
Explanation:
- Since, the no. of millimoles of NH₃ is more than that of HNO₃, the medium will be basic.
C of base (NH₃) = [(MV)NH₃ - (MV)HNO₃] / Vtotal.
C of base (NH₃) = [(0.20 M)(75.0 mL) - (0.50 M)(19.0 mL)] / (94.0 mL) = 0.0585 M.
∵ [OH⁻] = √(Kb.C)
∴ [OH⁻] = √(1.8 x 10⁻⁵)(0.0585 M) = 1.053 x 10⁻⁶.
∵ pOH = - log[OH⁻] = - log (1.053 x 10⁻⁶) = 5.97.
∴ pH = 14 - pOH = 14 - 5.97 = 8.03.
