How do I calculate the #"pH"# of the buffer solution when #1.00# #"g"# of potassium ethanoate is dissolved in #50.0# #"cm"^3# of #0.20# #"mol"# #"dm"^-3# ethanoic acid (#"K"_"a"=1.74xx10^-5# #"mol"# #"dm"^-3#)?

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Answer 1 · 2017-04-19T15:38:25

#"pH"=4.77#

#rho#(potassium ethanoate) #=1/(50divide1000) =20# #"g"# #"dm"^-3#

#rho="c"xx"Mr"therefore#

[potassium ethanoate]#=20/(39.1+24+32+3)=0.20387#

#["H"^+]="K"_"a"xx[["HA"]]/[["A"^-]]=1.74xx10^-5xx0.2/0.20387=1.7xx10^-5#

#"pH"=-log(1.7xx10^-5)=4.77#