13) Calculate the pH: a) 25.00 mL of 0.17 M ethylamine is titrated with 0.20 M h

Question

13) Calculate the pH:
a) 25.00 mL of 0.17 M ethylamine is titrated with 0.20 M hydrochloric acid. When 13.55 mL of the acid have been added what should the approximate pH be?
b) If pKw at 10.0°C is 14.534, what is the pH of pure water at this temperature?

13) Calculate the pH:
a) 25.00 mL of 0.17 M ethylamine is titrated with 0.20 M hydrochloric acid. When 13.55 mL of the acid have been added what should the approximate pH be?
b) If pKw at 10.0°C is 14.534, what is the pH of pure water at this temperature?

13) Calculate the pH:
a) 25.00 mL of 0.17 M ethylamine is titrated with 0.20 M hydrochloric acid. When 13.55 mL of the acid have been added what should the approximate pH be?
b) If pKw at 10.0°C is 14.534, what is the pH of pure water at this temperature?
b) If pKw at 10.0°C is 14.534, what is the pH of pure water at this temperature?

Explanation / Answer

a) ethyl amine has pkb = 3.2 ( taken from web)

we have reaction C2H5NH2 (aq) + H+ (aq) <---> C2H5NH3+ (aq)

C2H5NH2 initial moles = M x V (inL) = 0.17 x 0.025 = 0.00425

H+ moles = HCl moles = 0.2 x 0.01355 = 0.00271

C2H5NH2 moles left after reatcion with H+ = 0.00425-0.00271 = 0.00154

moles of C2H5NH3+ = H+ moles reacted = 0.00271

we use Henderson eq to find pH

pOH = pkb + log ( conjugate acid moles / base moles) where C2H5NH2 is base , C2H5NH3+ is conjugate acid

pOH = 3.2 + log ( 0.00271/0.00154)

= 3.446

pH = 14- 3.446 = 10.55

b) pkw = 14.534

pH = 1/2 ( pkw ) = 1/2 ( 14.534) = 7.267

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