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
