Just as pH is the negative logarithm of [H3O + ], p Ka is the negative logarithm

Question

Just as pH is the negative logarithm of [H3O + ], p Ka is the negative logarithm of Ka, p Ka = - log Ka The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: pH=p Ka + log [base]/[acid] Notice that the pH of a buffer has a value close to the p Ka of the acid, differing only by the logarithm of the concentration ratio [base]/[acid]. How many grams of dry N H 4C1 need to be added to 1.60L of a 0.400 M solution of ammonia, to prepare a buffer solution that has a pH of 8.57? Kb for ammonia is 1.8 Times 10 - 5 Express your answer with the appropriate units.

Explanation / Answer

pH=14-(pKb+log(salt/acid))

=14+log(1.8*10^-5)-log(x/(1.6*0.4))

or x=3.1 moles

mass=3.1*53.5

=165.85 grams

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