Question

An analytical chemist is titrating 99.2 mL of a 0.6300 M solution of dimethylamine ((CH3 NH) with a 0.2500 M solution of HNO3. The p K, of dimethylamine is 3.27. Calculate the pH of the base solution after the chemist has added 213.6 mL of the HNO3 solution to it.

Answer 1

pH of the base solution after the chemist has added 213.6 mL of the HNO3 solution to it is 9.96

Step-by-step explanation:

Calculating the moles of dimethylamine

Moles of dimethylamine = molairty * volume in L = 0.63*0.0992 = 0.0625moles

Moles of HNO3 = molarity * volume in L = 0.25*0.2136 = 0.0534 moles

The balanced chemical equation is

(CH3)2NH + HNO3 -------------------> (CH3)2NH2^+ + NO3^-

I 0.0625 0.0534 0

C -0.0534 - 0.0534 0.0534

E 0.0091 0 0.0534

As we know ,

POH = Pkb + log[(CH3)2NH2^+]/[(CH3)2NH]

= 3.27 + log0.0534/0.0091

= 3.27+ 0.768

= 4.038

PH = 14-POH

= 14-4.038 = 9.96