In: Chemistry
Calculate the pH of the solution when the following substances are added together
200 mL of 0.1 M NaOH and 80 mL of 2.5 M Acetic Acid
200 mL of 0.1 M NaOH and 8 mL of 0.25M Acetic Acid
no of moles of NaOH = molarity* volume in L
= 0.1*0.2 = 0.02moles
no of moles of CH3COOH = molarity * volume in L
= 2.5*0.08 = 0.2moles
CH3COOH + NaOH ----------> CH3COONa + H2O
I 0.2 0.02 0
C -0.02 -0.02 0.02
E 0.18 0 0.02
PH = PKa + log[CH3COONa]/[CH3COOH]
= 4.75 + log0.02/0.18
2.
no of moles of NaOH = molarity* volume in L
= 0.1*0.2 = 0.02moles
no of moles of CH3COOH = molarity * volume in L
= 0.25*0.008 = 0.002moles
CH3COOH + NaOH ----------> CH3COONa + H2O
I 0.002 0.02 0
C -0.002 -0.002 0.002
remaining no of moles of NaOH = 0.02-0.002 = 0.018 moles
molarity of OH- = no of moles/total volume in L
= 0.018/0.208 = 0.0865M
POH = -log[OH-]
= -log0.0865 = 1.0629
PH = 14-POH
= 14-1.0629 = 12.9371