What is the pH solution containing 0.30 M HCOOH and 0.52 M HCOOK?Compare your result with the pH of a 0.30 M HCOOH solution
HCOOH(aq) + H2O(l) = H3O+(aq) + HCOO-(aq)
pKa = 3.77
pH=pKa+log([HCOO−][HCOOH])pH = pKa + log ( \frac{[HCOO^-]}{[HCOOH]} )pH=pKa+log([HCOOH][HCOO−])
pH=3.77+log([0.52][0.30])pH = 3.77 + log ( \frac{[0.52]}{[0.30]} )pH=3.77+log([0.30][0.52])
pH = 4.01
Ka=[HCOO−][H+][HCOOH]=x20.3Ka=\frac{[HCOO^-][H^+]}{[HCOOH]}=\frac{x^2}{0.3}Ka=[HCOOH][HCOO−][H+]=0.3x2
pKa=logx2−log0.3pKa=logx^2-log0.3pKa=logx2−log0.3
logx2=0.52+3.77=4.26logx^2=0.52+3.77=4.26logx2=0.52+3.77=4.26
logx=pH=4.26/2=2.13
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