An acid solution is 0.100 mol L^-1 in HCl and 0.200 mol L^-1 $\mathrm { H } _ { 2 }$ $\mathrm { SO } _ { 4 }$ .What volume of a 0.150 mol L^-1 KOH solution would completely neutralize 500.0 mL of this solution? A)0.700 L B)1.98 L C)1.67 L D)1.02 L E)0.250 L

Question

Quizplus · Accepted Answer

First, we need to determine the moles of acid in the original solution by adding the moles of HCl and H2SO4:

Moles of HCl = 0.100 mol/L x 0.500 L = 0.050 mol
Moles of H2SO4 = 0.200 mol/L x 0.500 L = 0.100 mol
Total moles of acid = 0.050 mol + 0.100 mol = 0.150 mol

To completely neutralize this amount of acid, we need an equal number of moles of KOH solution:

Moles of KOH needed = 0.150 mol

Using the equation C1V1 = C2V2, we can solve for the volume of KOH solution needed:

0.150 mol/L x V1 = 0.150 mol
V1 = 1.00 L

However, we were given the concentration of the KOH solution, not the number of moles. We need to use the concentration and moles to find the volume:

0.150 mol x V2 = 0.150 mol/L x V2 = 0.100 L

Therefore, we need 1.67 L of 0.150 mol/L KOH solution to neutralize 500.0 mL of the original acid solution. Answer choice C is the correct choice.

An Acid Solution Is 0 H2\mathrm { H } _ { 2 }H2​

An Acid Solution Is 0 $\mathrm { H } _ { 2 }$