Calculate the volume of a 0.0321 M NaOH that will be required to neutralize 25.00mL of a 0.0399M HCL solution?

Question

Calculate the volume of a 0.0321 M NaOH that will be required to neutralize 25.00mL of a 0.0399M HCL solution?

Answer 1

HCl + NaOH --> H₂O + Na⁺ + Cl^_-

 

So we can use the 1:1 mole ratio to calculate the volume of the 0.0321 M NaOH needed to neutralize the solution.

 

Calculate the moles of HCl: (0.0399 mol/L)(0.025L) = 9.975 x 10^-4 mol HCl

 

Because its a 1:1 ratio the moles of HCl are equal to the moles of NaOH at the endpoint (neutralized solution), so we can then find the volume (V) of the NaOH solution

 

0.0321 M = 0.0321 mol/L = 9.975 x 10^-4 mol/V

V = 9.975 x 10^-4 mol/(0.0321 mol/L) --> V will be in liters