Question

An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of the acidic solution to completely neutralize all of the acid?

Options:
A) 300.0 mL
B) 250.0 mL
C) 200.0 mL
D) 150.0 mL

Answer 1

To neutralize all of the acid, we need to determine the volume of a 0.150 M solution of KOH that must be added to the acidic solution. Using stoichiometry, we can calculate the amount of KOH needed and then use the concentration of the KOH solution to calculate the volume of the KOH solution needed. The correct answer is option A) 300.0 mL.

Step-by-step explanation:

To completely neutralize all of the acid, we need to determine the volume of a 0.150 M solution of KOH that must be added to the acidic solution. Since HCl is in excess, we only need to consider the amount of H2SO4. From the balanced chemical equation, we know that 1 mole of H2SO4 reacts with 2 moles of KOH. Using stoichiometry, we can calculate the amount of KOH needed to neutralize the H2SO4. Given that the concentration is 0.210 M and the volume is 500.0 mL, we can calculate the amount of H2SO4 in moles by multiplying the concentration by the volume and dividing by 1000. This gives us 0.105 moles of H2SO4. Since the stoichiometric ratio is 2:1, we need double the amount of KOH. Therefore, the amount of KOH needed is 0.210 moles. Now, using the concentration of the KOH solution and the amount of KOH needed, we can calculate the volume of the KOH solution by dividing the moles of KOH by the concentration and multiplying by 1000. This gives us a volume of 1600 mL, which is equal to 1600 mL - 500 mL, = 1100 mL more than is required. Therefore, the correct answer is option A) 300.0 mL.