The hydronium ion concentration of an aqueous solution of 0.502 M pyridine (a weak base with the formula C5H5N) is

1.99×10¯¹⁴ M

Explanation:

We'll begin by writing the balanced dissociation equation of pyridine. This is illustrated below:

C₅H₅N + H₂O <=> C₅H₆N⁺ + OH¯

From the balanced equation above,

1 mole of C₅H₅N produced 1 mole of OH¯.

Therefore, 0.502 M C₅H₅N will also produce 0.502 M OH¯.

Finally, we shall determine the concentration of hydronium ion, H₃O⁺ in the solution. This can be obtained as follow:

Concentration of Hydroxide ion [OH¯] = 0.502 M

Concentration of hydronium ion [H₃O⁺] =?

[H₃O⁺] [OH¯] = 1×10¯¹⁴

[H₃O⁺] × 0.502 = 1×10¯¹⁴

Divide both side by 0.502

[H₃O⁺] = 1×10¯¹⁴ / 0.502

[H₃O⁺] = 1.99×10¯¹⁴ M

Thus, the concentration of hydronium ion, H₃O⁺ in the solution 1.99×10¯¹⁴ M

they are all solids.