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Considering the $K_\text{a}$ of the weak acid and the $K_\text{b}$ of the weak base that form the salt, we conclude that the solution will be acidic if $K_\text{b}$ > $K_\text{a}$ and basic if $K_\text{a}$ > $K_\text{b}$

I was wondering if anyone could clarify what the book is saying. My question is If we mix a solution with a $K_\text{b}$ of 0.1 with a solution with a $K_\text{a}$ of 0.001 would the final solution be acidic or basic?

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If there is a solution of mix of a weak acid/base and it's alkaline/acidic salt in molar ratio 1:1, then

  • The solution is acidic, if $K_\mathrm{a} \gt K_\mathrm{b}$, like acetic acid + sodium acetate(alkaline base)

  • The solution is alkalic(ine)/basic, if $K_\mathrm{a} \lt K_\mathrm{b}$, like ammonium chloride ( acidic salt) + ammonia(base)

Note that $$\begin{align} K_\mathrm{w}&=\ce{[H+][OH-]} \\ K_\mathrm{a}&=\frac{\ce{[H+][A-]}} {\ce{[HA]}} \\ K_\mathrm{b}&=\frac{\ce{[HA][OH-]}} {\ce{[A-]}} \\ &=\frac{\ce{[HA] K_\mathrm{w}}}{\ce{[A-][H+]}} \\ &=\frac{\ce{ K_\mathrm{w}}}{K_\mathrm{a}} \\ K_\mathrm{w}&=K_\mathrm{a} \cdot K_\mathrm{b} \\ \end{align}$$

Similarly, if there is such a 1:1 (molar amount) solution of an acid and a base, that are not in the mutual relation acid/salt or base/salt, the above is valid as well.

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Ka and kb are related as - Kw = ka × kb Here, kw is equlibrium constant of self ionisation of water. At a particular temperature kw is constant. E.g value of kw at 25 °C is 10^(-14). So, ka and kb are inversely propotional . Acidic and basic strength is inversely proportional to ka and kb respectively bescause measure of its strength is negative of log(ka or kb).

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  • $\begingroup$ Thank you for your answer. If we mix a solution with a Kb of 0.1 with a solution with a Ka of 0.001 would the final solution be acidic or basic $\endgroup$ – John May 1 at 12:43
  • $\begingroup$ That is s wrong question. Ka is a constant of acidic form of a compound, independently in if it exists as an acid or is neutralised, becoming a basic salt , or can be in any middle step. $\endgroup$ – Poutnik May 1 at 12:49

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