Imagine that you are in chemistry lab and need to make 1.00L of a solution with
ID: 795689 • Letter: I
Question
Imagine that you are in chemistry lab and need to make 1.00L of a solution with a pH of 2.40.
You have in front of you:
100mL of 6.00x10^-2M HCl
100mL of 5.00x10^02M NaoH
Plenty of distilled water
You start to add HCl to the beaker of water when someone asks you a quesitons. when you return to you dilution you accidently grab the wrong cylinder and add some NaOH. Once you realize your error you assess the situation. you have 85.0mL of HCl and 89.0mL of NaOH left in the original containers.
Assuming the final solution will be diluted to 1.00L, how much more HCl should yo uadd to acheive the desired pH?
Explanation / Answer
Final concentration of H+ = 10^(-pH) = 10^(-2.40) = 3.981 x 10^(-3) M
Final moles of H+ = final volume x final concentration
= 1.00 x 3.981 x 10^(-3) = 3.981 x 10^(-3) mol
Volume of HCl added = 100 - 85.0 = 15.0 mL
Moles of HCl added = 15.0/1000 x 6.00 x 10^(-2) = 9 x 10^(-4) mol
Volume of NaOH added = 100 - 89.0 = 11.0 mL
Moles of NaOH added = 11.0/1000 x 5.00 x 10^(-2) = 5.5 x 10^(-4) mol
NaOH + HCl => NaCl + H2O
Moles of H+ already present = excess moles of HCl = 9 x 10^(-4) - 5.5 x 10^(-4)
= 3.5 x 10^(-4) mol
Additional moles of H+ required = final moles of H+ - moles of H+ already present
= 3.981 x 10^(-3) - 3.5 x 10^(-4) = 3.631 x 10^(-3) mol
Volume of HCl required = additional moles of H+/concentration of HCl
= 3.631 x 10^(-3)/6.00 x 10^(-2)
= 0.0605 L = 60.5 mL