The composition of an acidic buffer made up of HA and NaA of total molarity 0.29 having pH =4.4 and Ka=1.8×10^-5 in terms of concentration of salt and acid respectively is

Dear Student,

The computation is expressed below,


pH = -log Ka + log [Salt][Acid]by considering a mollt for salt abd b =0.29-a for acid we get,4.4 = -log (1.8×10-5) + log a0.29-aa = 0.09[Acid] = 0.29-a = 0.29-0.09 = 0.20M[Salt] = a = 0.09 M

Regards.

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