PICTURE ATTACHED. ANSWER PART F ONLY!!! Consider the titration of a 24.0 ?mL sam

Question

PICTURE ATTACHED. ANSWER PART F ONLY!!!

Consider the titration of a 24.0 ?mL sample of 0.105 M HC2H3O2 with 0.130 M NaOH. Determine each of the following.

Part A - the initial pH (answered in picture)

Part B - the volume of added base required to reach the equivalence point (answered in picture)

Part C - the pH at 6.00 mL of added base (answered in picture)

Part D - the pH at one-half of the equivalence point (answered in picture)

Part E - the pH at the equivalence point (answered in picture)

Part F - the pH after adding 5.00 mL of base beyond the equivalence point

pH 2.86 2.52 v X 0.130 of added 0.05 8 30 log o 02,6 39 D) At are equivalerne point Salt Add PH pka 4.74 PH 2, 5 2. 2.4 -ma 4 8,75 1.52 At 3, 4 0,058M

Explanation / Answer

(f)

NaOH volume = after equivalence point = 24 + 5=29 ml

[NaOH] milli moles = 0.130 x29= 3.77 mmoles

here [HBr] <[KOH]

[OH-]= (3.77 - 2.52)/(29+24)

         = 0.0274 M

pOH= -log([OH])=1.627

pH= 14-pOH

   pH = 12.37

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