# Chemistry

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A titration is performed by adding .600 M KOH to 40.0 mL of .800 M HCl.

Calculate the pH before addition of any KOH.
Calculate the pH after the addition of 5.0 mL of the base.
Calculate the volume of base needed to reach the equivalence point.
What is the pH at the equivalence point (I say 7 because they're both Strong Acid/Strong Base, but not sure).
Calculate the pH after adding 5.00 mL of KOH past the equivalence point.

• Chemistry - ,

a)HCl --> H^+ + Cl^-
HCl is 100% ionized; therefore, (HCl) = (H^+) = 0.8M
pH = -log(H^+) = ??

b)to start. moles HCl = M x L = 0.800*0.040 = 0.032.
moles KOH added in 5 mL = M x L = 0.600*0.005 = 0.003
................HCl + KOH ==> KCl + H2O
initial......0.032....0........0.....0
change...-0.003...-0.003..+0.003.+0.003
equil.....0.029.....0.....0.003.0.003
So you essentially have 0.029 moles HCl in 45 mL (0.045L) so (HCl) = )H^+) = 0.029/0.045 = 0.644 and pH = -log(H^+) = ??

c) moles HCl to start = M x L = ??
moles KOH needed = same as moles HCl
M KOH = moles KOH/L KOH

d)yes, NaCl is the salt of a strong acid and strong base; therefore, the pH at the equivalence point is 7

e) Do this the same way as the (b) part (the addition of 5 mL KOH) EXCEPT here you are 5 mL past the equivalence point and the principle species in solution is excess KOH. Calculate pOH and pH from that.