(show all your works for all questions) 1. a 30.0ml sample of 0.030M acetic acid
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(show all your works for all questions)
1. a 30.0ml sample of 0.030M acetic acid is titrated with 0.025M potassium hydroxide. calculate the pH after the following volumes of base have been added:
(i)00.0ml
(ii)18.0ml
(iii)36.0ml
(iv)45.0ml
2. use the picture below, which indicator, phenolphthalein or bromthymol blue, is likely to be the better choice for this titration?why?
pH 7 2 4 8 10 11 12 1314 Crystal violet Thymol blue 4Dinitrophenol Bamophenol blue omacresol green Methyl red Alizarin onothymol blue Phenol red Phenolphthaleir Alzarin yellow R gure 19.6 Colors and 136 Colors and approximate pH ranges of some common acid-base indicators.Explanation / Answer
Acetic acid is weak mono basic acid, KOH also mono acidic strong base.
therefore at end point of the titration of above solutions,
no. of moles of base = no. of moles of acid
no. of moles of acid = molarity X volume (L) = 0.03 X 0.03 = 0.0009 moles (or) 0.9 mmoles.
at various base conditions: KOH concentration = 0.025 M
(i) at its 0 mL volume
no. of base moles = M X V (mL) = 0 mmol.
no overall volume of solution increses.
So, the acid concentration oes not alter.
pH = -log [H+] = -log 0.03 = 1.52
(ii) at 18 mL of base
therefore no. of base moles = M X V (mL) = 0.025 X 18 = 0.45 mmol.
overall volume of solution = 30 + 18 = 48 mL
no. of acid moles remaing after titration with 18 mL of NaOH = 0.45 mmol
remaining acid concentration = mmol/ml = 0.45/48 = 0.0094 M
pH = -log [H+] = -log 0.0094 = 2.03
(iii) at 36 mL of base
therefore no. of base moles = M X V (mL) = 0.025 X 36 = 0.9 mmol.
overall volume of solution = 30 + 36 = 66 mL
no. of acid moles remaing after titration with 18 mL of NaOH = 0 mmol
this is end point of titration, at this point pH = 7
(iv) at 45 mL of base
therefore no. of base moles = M X V (mL) = 0.025 X 45 = 1.125 mmol.
overall volume of solution = 30 + 45 = 75 mL
base moles are more than acid moles
no. of base moles remaing after titration = 0.225 mmol
remaining base concentration = mmol/ml = 0.225/75 = 0.003 M
pOH = -log [OH-] = -log 0.003 = 2.52
pH = 14 - pOH = 11.48
2. This is the titration between weak acid and strong base. At its end point pH of solution shifts from about 6 to 10 (approximately) so, the indicator change in this region is useful for this titration.
Phenolpthalein is the better suited one for this.
Best wishes.
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