A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.3

Question

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+.

Explanation / Answer

HCl + NaOH => NaCl + H2O

Moles of NaOH = moles of HCl = volume x conentration of HCl

= 503.4/1000 x 0.315 = 0.158571 mol


H2SO4 + 2 NaOH => Na2SO4 + 2 H2O

Moles of H2SO4 = volume x concentration of H2SO4

= 503.4/1000 x 0.125 = 0.158571 mol


Moles of NaOH = 2 x moles of H2SO4

= 2 x 0.158571 = 0.12585 mol


Total moles of NaOH = 0.158571 + 0.12585 = 0.284421 mol


Volume of NaOH = moles/concentration of NaOH

= 0.284421/0.55

= 0.5171 L = 517.1 mL

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