log

K'_p

K_p

= $$ - {{\Delta H} \over {2.303R}}\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]$$

For exothermic reaction, $$\Delta $$H = -ve means the temperature T₂ is higher than T₁.

$$ \therefore $$ $$\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]$$ is negative.

So log K'_p - log K_p = -ve

$$ \Rightarrow $$ log K_p > log K'_p

$$ \Rightarrow $$ K_p > K'_p

We know, $$\Delta $$G^o = – 2.303 RT log K_sp

$$ \therefore $$ 63300 = – 2.303 × 8.314 × 298 log K_sp

$$ \Rightarrow $$ log K_sp = -11.09

$$ \Rightarrow $$ K_sp = 10^-11.09

= 8.0 × 10^–12

Na₂CO₃ is salt of strong base, NaOH and weak acid, H₂CO₃ hence the pH value of the solution will be high.

Given, K_a = 1 × 10^–4

pK_a = – log (1× 10^–4) = 4

pH = pK_a + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ 5 = 4 + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 1

$$ \Rightarrow $$ $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 10 = 10 : 1