# Problem: Given a weak acid HA and the following equilibrium in water:HA(aq) ⇌ H+(aq) + A–(aq)which answer describes a situation where Kb for the conjugate base A– will have the largest Kb value?A. Ka = 7.0 × 10–8   and   T = 25 °CB. Ka = 7.0 × 10–8   and   T = 80 °CC. Ka = 9.0 × 10–6   and   T = 25 °CD. Ka = 9.0 × 10–6   and   T = 80 °C

###### FREE Expert Solution

We are being asked to find determine which situation where Kb for the conjugate base A will have the largest Kb value.

HA(aq)            H3O+(aq)  +    A-(aq)
(weak acid)                                        (conjugate base)

Hence, the Kb of A- can be calculated from the Ka of HA.  Ka and Kb are connected by the autoionization constant of water (Kw) in the following equation:

$\overline{){{\mathbf{K}}}_{{\mathbf{w}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{·}}{{\mathbf{K}}}_{{\mathbf{b}}}}$

Kw = 1.0x10-14             @ 25°C

${\mathbf{K}}_{{\mathbf{w}}}\mathbf{=}{\mathbf{K}}_{{\mathbf{a}}}\mathbf{·}{\mathbf{K}}_{{\mathbf{b}}}\phantom{\rule{0ex}{0ex}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}}}\mathbf{=}\frac{\overline{){\mathbf{K}}_{\mathbf{a}}}\mathbf{·}{\mathbf{K}}_{\mathbf{b}}}{\overline{){\mathbf{K}}_{\mathbf{a}}}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}}}}$ ###### Problem Details

Given a weak acid HA and the following equilibrium in water:

HA(aq) ⇌ H+(aq) + A(aq)

which answer describes a situation where Kb for the conjugate base A will have the largest Kb value?

A. Ka = 7.0 × 10–8   and   T = 25 °C

B. Ka = 7.0 × 10–8   and   T = 80 °C

C. Ka = 9.0 × 10–6   and   T = 25 °C

D. Ka = 9.0 × 10–6   and   T = 80 °C