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

1 answer

1. 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.

   Answer ID
   524128

   Created
   April 2, 2011 2:30am UTC

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   https://questions.llc/answers/524128