Below is a titration curve for Lysine. If the titration is done by adding NaOH t
ID: 818658 • Letter: B
Question
Below is a titration curve for Lysine. If the titration is done by adding NaOH to a solution containing 4.098 moles of lysine, how many milliliters (mL) of a 10 M solution of NaOH would you have to add to reach point "C" equivalent? Enter your answer in mL with 2 decimal points. Assume that the amount of NaOH added does not have a significant effect on the final volume of the buffer. How many mL of NaOH
Below is a titration curve for Lysine. If the titration is done by adding NaOH to a solution containing 4.098 moles of lysine, how many milliliters (mL) of a 10 M solution of NaOH would you have to add to reach point "C" equivalent? Enter your answer in mL with 2 decimal points. Assume that the amount of NaOH added does not have a significant effect on the final volume of the buffer. How many mL of NaOHExplanation / Answer
Lysine is effective a tripotic acid with 3 pKa values
=> it can be represented as H3A
Point A = 1st equivalence point:
H3A + NaOH => H2A- + Na+ + H2O
Moles of NaOH added = moles of H3A = 4.098 mol
Volume of NaOH added = moles/concentration of NaOH
= 4.098/10 = 0.4098 L = 409.80 mL
Point D = 2nd equivalence point
H2A- + NaOH => HA2- + Na+ + H2O
Additional Moles of NaOH added = moles of H2A- = 4.098 mol
Additional Volume of NaOH added = moles/concentration of NaOH
= 4.098/10 = 0.4098 L = 409.80 mL
Point C = halfway to 2nd equivalence point where [H2A-] = [HA2-] and pH = pKa2
Volume of NaOH = volume for 1st equivalence point + (1/2) x additional volume for 2nd equivalance point
= 409.80 + 1/2 x 409.80
= 614.70 mL
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