Question

25 ml of 0.4M a weak base and 75 ml of 0.2 M of its salt forms a buffer solution. K of weak base is 2 x 10 5. The pH of the buffer solution is log 1.5 = 0.176
help ​

Answer 1

pH of basic buffer is-
pH=14-pKb-log (salt/base)
=14+log Kb -log (salt/base)
=14+log (2*10^-(5)) -log ((0.2*75/1000)/(0.4*25/1000))
=14+log 2 -5- log 1.5
=14+0.3-5-0.176=9.124=9.12 approx