7. In addition to its pharmaceutical uses, barbituric acid (HC4H3N2O3; abbreviated as HBarb; Ka = 9.8 x 10-5) has uses in textiles and plastics. Suppose a chemist was going to perform a titration analysis on 25.0 mL of 0.20 M HBarb using 0.18 M NaOH. This titration would require 27.8 mL to reach the equivalence point. How many moles of NaOH remains at the end of the reaction after the addition 10.00 mL?
a) 0.0 x 10-0
b) 1.8 x 10-3
c) 5.0 x 10-3
d) 3.2 x 10-3
Correct ans- Option d
In a titration, Equivalence point is that point where all the acid presentin the solution have completely been neutralized by the base added. On further addition of base after this point only make the solution basic as there is no more acid present for neutralization.
Now given the weak acid HC4H3N2O3 taken has Ka = 9.8 x 10-5
Now when we add a strong base like NaOH, then it react with the acid to from the salt of its conjugate base i.e
NaOH + HC4H3N2O3 --------------> NaC4H3N2O3 + H2O
It shows that 1 mole of NaOH is required to neutralize 1 mole of HC4H3N2O3
Now given HC4H3N2O3 taken = 25.0 mL of 0.20 M
That means mols of HC4H3N2O3 taken = concentration * volume
= 0.20 M * 25.0 mL
= 0.20 mol/1000 ml * 25.0 mL
= 0.005 mols
That means total mols of NaOH required to neutralize this = 0.005 mols
Now given titration would require 27.8 mL to reach the equivalence point.
That means in our 27.8 mL NaOH solution, we have 0.005 mols NaOH
Then in 10 ml of NaOH we have = 0.005 mols/ 27.8 mL NaOH * 10 ml of NaOH
= 0.00179 mols NaOH
So after addition of 10 ml addition, the rest of NaOH left = 0.005 mols - 0.00179 mols
= 0.00321 mols
= 3.2 * 10-3 mols
7. In addition to its pharmaceutical uses, barbituric acid (HC4H3N2O3; abbreviated as HBarb; Ka = 9.8...
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