This can be calculated using #K_w# for water at 298K .
#K_w# for water at 298K is #1xx10^-14# .
#K_w = K_a "of acid" xxK_b "of conjugate base"#
#pK_w = pK_a "of acid" xx pK_b "of conjugate base"#
#K_w = K_a "of base" xxK_b "of acid"#
#pK_w = pK_a "of base" xx pK_b "of conjugate acid"#
#K_w = OH^-)"conc" xx H^+"conc"#
At different temperatures Kw has different values
#K_w# is also converted to #pK_w #.
#pK_w = -log_10(K_w)#
10 is the base
First write down the dissociation of benzoic acid.
#C_6H_6COO + H_2O= C_6H_5COO + H_3O^+#
Recall the above rules
#K_w = K_a "of acid" xxK_b "of conjugate base"#
#1xx10^-14 = K_a "of benzoic acid" xx K_b of C_6H_5COO^-#
#1xx10^-14 = 6.3xx10^-5 xx Kb "of C6H5COO"#
#K_b = (1xx10^-14)/(6.3xx10^-5)#
= #1.5873xx10^-10#
Again apply the same rule
Write down the dissociation of #HOCH_2CH_2NH_2#
#HOCH_2CH_2NH_2 = HOCH_2CH_2NH_3^+ + OH^-#
First we must obtain Kb from pKb
#10^(-pK_b) = K_b#
#10^-4.49 =0.00003 = 3*10^-5#
#K_w = KbxxKa#
#1 xx 10^-14 = 3*10^-5 xx x#
#x = (1xx10^-14)/(3*10^-5)#
#K_a = 3.33333*10^-10#
