For the titration designated 1 calculate the pH at the following percentages-of
ID: 881000 • Letter: F
Question
For the titration designated1 calculate the pH at the following percentages-of the stoichiometric amount of titrant: 0 (initial point), 10, 50, 90, 95, 99, (stoichiometric point), 101, 105, 110. Construct the titration curve by plotting pH against percent titration. State whether or not the titration is feasible; if feasible, select an appropriate indicator.
Solution to be titrated Titrant
(a)50.0 mL of 0.0100 N HCl 0.0100 N NaOH
(b) 50.0 mL of 0.200 N HC2H3O2 0.200 N NaOH
(c) 25.0 mL of 0.250 N NH4OH 0.250 N HCl
Explanation / Answer
(a) Solution to be titrated - 50.0mL of 0.0100 N HCl.
Given normality, N of HCl = 0.0100 N
Since HCl is a monovalent compound, n-factor is 1 and normality(N) = molarity(M)
Hence [HCl] = 0.0100 M
Since HCl is a strong acid it is completely ionised. Hence
[H+] = [HCl] = 0.0100M
Hence pH at 0(initial point) = - log[H+] = - log(0.0100M) = 2 (answer)
The titrant used is 0.0100N NaOH or 0.0100 M NaOH
At equivalence point, all the acid must be neutralised by NaOH.
Hence at equivalence point moles of HCl = moles of NaOH added.
Let 'V' be the volume of NaOH added to achieve the equivaalence point.
Moles of HCl = M*V = 0.0100M*50.0mL*(1L/1000mL) = 0.00050 mol = 0.5 mmol
Moles of NaOH added to achieve equivalence point = M*V = 0.0100V
at equivalence point moles of HCl = moles of NaOH added.
.=> 0.00050 mol HCl = 0.0100 V NaOH
=>V = 0.00050 / 0.0100 L = 0.050 L NaOH = 50 mL NaOH.
10% of stoichiometric amount of titrant(NaOH) = (50 mL NaOH) * (10 / 100) = 5 mL NaOH.
Hence millimoles of NaOH added = 0.0100M*5mL = 0.05 millimol
HCl + NaOH ----> Na+ + Cl- + H2O
Init: 0.5mmol 0.05mmol 0
final: 0.45mmol 0 0.05 mmol
Hence 0.05 mmol of NaOH will be neutralized by 0.05 mmol of HCl, leaving 0.45mmol of excess HCl.
Hence [H+] = [HCl] = mmoles of HCl / Total volume(mL) = 0.45 mmol / (50+5)mL = 0.00818M
Hence pH = - log[H+] = - log(0.00818) = 2.09 (answer)
50% of stoichiometric amount of titrant(NaOH) = (50 mL NaOH) * (50 / 100) = 25 mL NaOH.
millimoles of NaOH added = 0.0100M*25mL = 0.250 millimol
Hence 0.25 mmol of NaOH will be neutralized by 0.25 mmol of HCl, leaving 0.25mmol of excess HCl.
[H+] = [HCl] = mmoles of HCl / Total volume(mL) = 0.25 mmol / (50+25)mL = 0.00333M
Hence pH = - log[H+] = - log(0.00333) = 2.48 (answer)
90% of stoichiometric amount of titrant(NaOH) = (50 mL NaOH) * (90 / 100) = 45 mL NaOH.
millimoles of NaOH added = 0.0100M*45mL = 0.450 millimol
Hence 0.45 mmol of NaOH will be neutralized by 0.45 mmol of HCl, leaving 0.05mmol of excess HCl.
[H+] = [HCl] = mmoles of HCl / Total volume(mL) = 0.05 mmol / (50+45)mL = 0.000526M
Hence pH = - log[H+] = - log(0.000526) = 3.28 (answer)
99% of stoichiometric amount of titrant(NaOH) = (50 mL NaOH) * (99 / 100) = 49.5 mL NaOH.
millimoles of NaOH added = 0.0100M*49.5mL = 0.495 millimol
Hence 0.495 mmol of NaOH will be neutralized by 0.495 mmol of HCl, leaving 0.005mmol of excess HCl.
[H+] = [HCl] = mmoles of HCl / Total volume(mL) = 0.005 mmol / (50+49.5)mL = 0.0000502M
Hence pH = - log[H+] = - log(0.0000502) = 4.30 (answer)
105% of stoichiometric amount of titrant(NaOH) = (50 mL NaOH) * (105 / 100) = 52.5 mL NaOH.
millimoles of NaOH added = 0.0100M*52.5mL = 0.525 millimol
Hence 0.525 mmol of NaOH will be neutralized by 0.500 mmol of HCl, leaving 0.025mmol of excess HaOH making the solution basic.
[OH-] = [NaOH] = mmoles of NaOH / Total volume(mL) = 0.025 mmol / (50+52.5)mL = 0.000244M
Hence pOH = - log[OH-] = - log(0.000244) = 3.61
=> pOH = 14 - 3.61 = 10.39 (answer)
Similarly we can calculate for 110%.
This titration is feasible and we can use phenolphthalein as the indicator, because we can easily observe the physical change caused by phenolphthalein during pH change.
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