Part C A 75.0-mL volume of 0.200 mol L−1 NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 mol...

Part C

A 75.0-mL volume of 0.200 mol L−1 NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 mol L−1 HNO3. Calculate the pH after the addition of 28.0 mL of HNO3.

Express your answer numerically.

Part D

A 52.0 mL volume of 0.350 mol L−1 CH3COOH (Ka=1.8×10−5Ka=1.8×10−5) is titrated with 0.400 mol L−1 NaOH. Calculate the pH after the addition of 15.0 mL of NaOH.

Express your answer numerically.

Solutions

Expert Solution

queen_honey_blossom answered 1 year ago

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