Posted by Summer on Friday, April 1, 2011 at 1:18pm.
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
added KOH.............0.003..........
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.
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