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concentration of reN solution =1xx10^(-2)M=0.01M HeN dissociates in sahes solution as follows.HCN⇌H^(+)+CN^(-)longrightarrow" (1) "Changeinlone -xM+xM+xMInitial cone. 0.01MEnd coneantation (0.01-x)M quad xM quad xMTherefore x moles of H^(+)and x moles of CN^(-) dissoliates from H+Cl and the Concentration of HeN remains (0.01-x)M.Now K_(b) for eN^(-)=1.6 xx10^(-5).Again we know k_(a)k_(b)=k_(w)=1xx10^(-14):.Ka=(1xx10^(-14))/(1.6 xx10^(-5))=6.25 xx10^(-10):. Kavalue for H+CN=6.25 xx10^(-10)From equat ... See the full answer