# Question 0837e

Mar 23, 2017

This can be calculated using ${K}_{w}$ for water at 298K .
${K}_{w}$ for water at 298K is $1 \times {10}^{-} 14$ .

${K}_{w} = {K}_{a} \text{of acid" xxK_b "of conjugate base}$

$p {K}_{w} = p {K}_{a} \text{of acid" xx pK_b "of conjugate base}$

${K}_{w} = {K}_{a} \text{of base" xxK_b "of acid}$

$p {K}_{w} = p {K}_{a} \text{of base" xx pK_b "of conjugate acid}$

K_w = OH^-)"conc" xx H^+"conc"#

At different temperatures Kw has different values

${K}_{w}$ is also converted to $p {K}_{w}$.

$p {K}_{w} = - {\log}_{10} \left({K}_{w}\right)$

10 is the base

First write down the dissociation of benzoic acid.

${C}_{6} {H}_{6} C O O + {H}_{2} O = {C}_{6} {H}_{5} C O O + {H}_{3} {O}^{+}$

Recall the above rules

${K}_{w} = {K}_{a} \text{of acid" xxK_b "of conjugate base}$

$1 \times {10}^{-} 14 = {K}_{a} \text{of benzoic acid} \times {K}_{b} o f {C}_{6} {H}_{5} C O {O}^{-}$

$1 \times {10}^{-} 14 = 6.3 \times {10}^{-} 5 \times K b \text{of C6H5COO}$

${K}_{b} = \frac{1 \times {10}^{-} 14}{6.3 \times {10}^{-} 5}$

= $1.5873 \times {10}^{-} 10$

Again apply the same rule

Write down the dissociation of $H O C {H}_{2} C {H}_{2} N {H}_{2}$

$H O C {H}_{2} C {H}_{2} N {H}_{2} = H O C {H}_{2} C {H}_{2} N {H}_{3}^{+} + O {H}^{-}$

First we must obtain Kb from pKb

${10}^{- p {K}_{b}} = {K}_{b}$

${10}^{-} 4.49 = 0.00003 = 3 \cdot {10}^{-} 5$

${K}_{w} = K b \times K a$

$1 \times {10}^{-} 14 = 3 \cdot {10}^{-} 5 \times x$

$x = \frac{1 \times {10}^{-} 14}{3 \cdot {10}^{-} 5}$

${K}_{a} = 3.33333 \cdot {10}^{-} 10$