Question

What is the concentration of hydronium ions after a 100mL solution of 0.02M Magnesium Hydroxide and 100mL of 0.05M nitric acid?

Answer 1

In this case, we must first determine the moles of H+ ions produced by the strong acid (HCl) and the moles of OH- ions produced by the strong base (Mg(OH)2) as form given reaction:

2HCl     +     Mg(OH)2   -------->     MgCl2     +      2H2O

Given,

For 100mL solution of 0.02M Magnesium Hydroxide:

N_baseV_base = (100 x 0.02) x 2 = 4.0 mol OH-

[ x 2 because, one mole of Mg(OH)2 will give 2 moles of OH-]

For 100mL of 0.05M nitric acid:

N_acidV_acid = 100 x 0.05 = 5.0 mol H+

As the moles of H+ are greater than the moles of OH-, the solution will be acidic in nature and hence, we must find the moles of excess H+ and finally the [H+] as:

N_resulting = [H3O+] = N_acidV_acid - N_baseV_base / V_acid + V_base

N_resulting = [H3O+] = 5.0 - 4.0 / (100 + 100)

N_resulting = [H3O+] = 1.0 / 200

N_resulting = [H3O+] = 0.005M (hydronium ions concentration)