Question

A 75.0-ml volume of 0.200 m nh3 (kb=1.8×10−5) is titrated with 0.500 m hno3. calculate the ph after the addition of 19.0 ml of hno3. express your answer numerically.

Answer 1

8.03.

Step-by-step 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.