In: Chemistry
calculate the volume (in milliliters) of 3.0M naoh that must be added to a 1.00M ch3cooh solution to make a buffer having ph of 4.90
PH = 4.9
PKa = 4.74
volume of CH3COOH = 1L
no of moles of CH3COOH = molarity * volume in L
= 1*1 = 1 moles
no of moles of NaOH = molarity * volume in L
= 3*v
---------- CH3COOH (aq) + NaOH(aq) ---------------> CH3COONa(aq) + H2O(l)
I--------------- 1 ------------------ 3v ---------------------------------- 0
C------------- -3v --------------- -3v ----------------------------------- 3v
E ---------- 1-3v ---------------- 0 ------------------------------------ 3v
PH = Pka + log[CH3COONa]/[CH3COOH]
4.9 = 4.74 + log[CH3COONa]/[CH3COOH]
log[CH3COONa]/[CH3COOH] = 4.9-4.74
log[CH3COONa]/[CH3COOH] =0.16
[CH3COONa]/[CH3COOH] = 1.4454
3v/(1-3v) = 1.4454
3v = 1.4454(1-3v)
v = 0.197L
volume of NaOH = 197ml >>>>answer