Is an aqueous solution of potassium bicarbonate acidic or basic in nature?

The following is the definition of an acid salt according to my textbook: The salts formed by the partial replacement of the replaceable hydrogen atoms (H+ ions) of an acid molecule by a metallic or ammonium ion are called acid salts. Acid salts ionise in water to give hydronium ions and therefore, their aqueous solutions show all the properties of an acid.

Going by this definition, potassium bicarbonate should be classified as an acid salt. Hence, an aqueous solution of potassium bicarbonate should be acidic in nature(as hydronium ions would be present in an aqueous solution of potassium bicarbonate).

However, my textbook also says, “The aqueous solutions of salts formed by a weak acid and strong base (as a result of a neutralisation reaction) are basic in nature.”

According to this, potassium bicarbonate solution should be classified as a solution which is basic in nature as potassium bicarbonate solution is formed as a result of a neutralisation reaction between potassium hydroxide, which is a strong base and concentrated carbonic acid, which is a weak acid even in its concentrated solution form.

So is potassium bicarbonate solution acidic or basic in nature?

• I see the term acidic is in the question context used in ambiguous way. acidic(1) as being an acid, acidic(2) as forming solution with pH<7. While bicarbonate solution is acidic(1), as HCO3- is a weak acid, it is not acidic(2), as HCO3- is also a not so weak base. – Poutnik Sep 1 '20 at 4:41
• Another example of an acid salt, that is not acidic(2) is Na2HPO4. The pH of an acid salt is approximately the average of 2 relevant neighbour pKas of respective acids. – Poutnik Sep 1 '20 at 4:43

You could simply calculate the $$\mathrm{pH}$$ of the solution. $$K_\mathrm{a{_1}}$$ of $$\ce{H2CO3}$$ is $$\pu{4.30E-7}$$ and $$K_\mathrm{a{_2}}$$ of $$\ce{H2CO3}$$ is $$\pu{4.72E-11}$$.The approximate $$\mathrm{pH}$$ of the solution can be given by: $$\mathrm{pH}=\frac{(\mathrm{p}K_\mathrm{a{_1}} + \mathrm{p}K_\mathrm{a{_2}})}{2}$$ This yields $$\mathrm{pH}= 8.347$$, which indicates that the solution is mildly basic.