A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.315 M HCl and 0.125 M H2SO4. What volume of 0.55 M NaOH would be required to completely neutralize all of the acid in 503.4 mL of this solution?

hints

involves a solution that has 2 different acids in it. One way to do this is to imagine that you are neutralizing 2 solutions, one of each acid, and then just add the amounts of base needed. Another way is to think about how many moles of H3O+ are present in the mixed acids, and then figure out how much of the basic solution is needed to react with that amount of H3O+.

The volume of NaOH required is 402.9 mL

Explanation:

To calculate the number of moles for given molarity, we use the equation:

    .....(1)

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

Hence, the volume of NaOH required is 402.9 mL

the answer would be b i believe