Calculate the hydronium ion concentration in an aqueous solution that contains 2.50 × 10-2 mol L-1 in hydroxide ion.

The concentration of hydronium ion in the solution is 3.98 × 10^-13 mol L-1.

From the question, we know that the concentration of hydroxide ions in the solution is 2.50 × 10-2 mol L-1.

pOH = - log[OH-]

pOH = -log[2.50 × 10-2 mol L-1]

pOH = 1.6

Recall that;

pOH + pH = 14

pH = 14 - pOH

pH = 14 - 1.6

pH = 12.4

But

pH = -log[H3O^+]

[H3O^+] = Antilog(-12.4)

[H3O^+] = 3.98 × 10^-13 mol L-1

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4×10¯¹³ molL¯¹

Explanation:

The following data were obtained from the question:

Concentration of Hydroxide ion [OH¯] = 2.5×10¯² molL¯¹

Concentration of hydronium ion [H₃O⁺] =.?

Thus, the concentration of the hydronium ion [H₃O⁺] can be obtained as follow:

[H₃O⁺] × [OH¯] = 1×10¯¹⁴ mol²L¯²

[H₃O⁺] × 2.5×10¯² = 1×10¯¹⁴

Divide both side by 2.5×10¯²

[H₃O⁺] = 1×10¯¹⁴ / 2.5×10¯²

[H₃O⁺] = 4×10¯¹³ molL¯¹

Therefore, the concentration of hydronium ion [H₃O⁺] in the aqueous solution is 4×10¯¹³ molL¯¹

the answer to your question is c.

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