Technically the concentrations go in in molarity. I like to work in mols and since M = mols/L and the value for L in both numerator and denominator is the same the L cancel and one can use mols (actually I use millimols) directly.
millimols HAc = 500 mL x 0.3M = 150 mmols.
........OH^- + HAc ==> Ac^- + H2O
I.......0......150......0......0
add.....x.......................
C......-x......-x.......x......
E.......0.....150-x.....x.......
Substitute into the HH equation and solve for x = mmols NaOH.
Then since M = mmols/mL, plug in M NaOH and mmols NaOH to find mL 0.422 M NaOH. I think the answer is approx 50 mL.
I....
what volume of .422 M NaOH must be added to .500L of .300 M acetic acid to raise its ph to 4.00
i know you have to use the henderson hasselbalch equation but im not sure which values i would use for the concentration portion of the equation. would i have to change the molarity of any of the amounts given initially.
would i add the naoh to the acetic acid which would go to completion leaving me with 0.078 M acetic acid and .422 M of the conjugate then use those values in the equation
.177827 = (.422)(x)/(.078)(.500-x)
then x would equal the volume of NaOh needed?
thank you in advance
1 answer
1 answer