What is the hydronium-ion concentration in a solution formed by combining 750 mL of 0.10 M NaOH with 250 mL of 0.30 M HCl?
NaOH(aq) + HCl(aq) $\to$ NaCl(aq) + H₂O(l)

Question

What is the hydronium-ion concentration in a solution formed by combining 750 mL of 0.10 M NaOH with 250 mL of 0.30 M HCl? NaOH(aq)+ HCl(aq)$\to$ NaCl(aq)+ H₂O(l)

Quizplus · Accepted Answer

First, we need to determine which reactant will be the limiting reactant.

Using stoichiometry, 0.30 moles of HCl reacts with 0.30 moles of NaOH to form 0.30 moles of NaCl.

0.10 moles of NaOH reacts with only 0.10 moles of HCl. Therefore, NaOH is limiting.

The amount of moles of NaOH used is (0.750 L)(0.10 mol/L) = 0.075 mol.

The total volume of the solution is 750 mL + 250 mL = 1000 mL = 1.00 L.

The concentration of hydronium ions can be found using the equation:

Kw = [H3O+][OH-] 
1.0 x 10^-14 = [H3O+][0.075/1.00]

[H3O+] = (1.0 x 10^-14)/(0.075/1.00) = 1.33 x 10^-13 M

Therefore, the answer is C, 1.0 x 10^-7 M.

What Is the Hydronium-Ion Concentration in a Solution Formed by Combining