Question

An acid solution is 0.110 M in HCl and 0.190 M in H₂SO₄.

What volume of a 0.160 M KOH solution would have to be addedto 500.0 mL of the acidic solution to neutralize completely all ofthe acid? (Answer must be in liters)

Answer 1

To neutralize the acid, 0.9375 liters (or 937.5 mL) of a 0.160 M KOH solution should be added to the 500.0 mL of the acidic solution.

Step-by-step explanation:

To neutralize all of the acid, we need to calculate the moles of HCl and H₂SO₄ present in the acidic solution, and then determine the moles of OH- required to neutralize those acids.

Moles of HCl: 0.500 L x 0.110 M = 0.055 moles

Moles of H₂SO₄: 0.500 L x 0.190 M = 0.095 moles

To neutralize these acids, we need an equal number of moles of OH-. Since KOH is monoprotic, each mole of KOH will react with one mole of acid.

Therefore, the total moles of OH- required is 0.055 moles + 0.095 moles = 0.150 moles.

To calculate the volume of a 0.160 M KOH solution needed to provide 0.150 moles of OH-, we can use the equation:

moles = volume (L) x concentration (M)

0.150 moles = volume (L) x 0.160 M

Volume (L) = 0.150 moles / 0.160 M = 0.9375 L.

Therefore, 0.9375 liters (or 937.5 mL) of the 0.160 M KOH solution would need to be added to neutralize all of the acid in the 500.0 mL acidic solution.