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+.

Answer 1

Answer: The volume of NaOH required is 402.9 mL

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$     .....(1)

• For HCl:

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:

$0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol$

• For sulfuric acid:

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

$0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol$

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:

$0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL$

Hence, the volume of NaOH required is 402.9 mL