Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

I\'m trying to find the Volume, PH and PKa for (1/4) equivalence point (1/2) equ

ID: 629252 • Letter: I

Question

I'm trying to find the Volume, PH and PKa for (1/4) equivalence point (1/2) equivalence point and at the 3/4 equivalence point. Molarity of NaOH used is 0.05131. 10.0 mL of unknown added to a flask with aprox 40 ml of water. After i titrated the weak acid with the strong base i plotted my point and found the equivalence point to be V=7.82 mL and a ph of 8. SO moles of acid is = to moles of base at equivalence point so moles was determined to be 0.0004012 moles and the molarity of unknown is 0.04012 M unknow acid. I know at 1/2 equivalence point ph=pka bc log 1 =1 for the henderson haselbach equation. But i'm stuck finding the pka and everything else. Please show all work

Explanation / Answer

The unknown acid HA reacts with NaOH in the following way


HA + NaOH ---> NaA + H2O


using M1V1 = M2V2 => [HA] = 0.05131*7.82/50 = 0.00802



mole of NaOH added to reach the equivalence point = 0.0004012, total volume = 50+7.82 = 57.82


hence [NaA] formed = 0.0004012/0.05782 = 0.0064



At equivalance point the pH is determined by the following reaction


A- + H2O ----> HA + OH-


given pH = 8 => [OH-] = 10^-6


now Kb = Kw/Ka = [HA][OH-]/[A-] => Kw/Ka = 10^-12/0.0064 => Kw/Ka = 1.5625 * 10^-10


=> Ka = 10^-14/1.5625* 10^-10 = 6.4 * 10^-5


Hende Ka = 6.4 *10^-5


Now Volume for 1/4 equivalence point = 7.82/4 = 1.955 mL


Hence mole of NaOH added = 1*10^-4


initial mole of HA = 0.00802*0.05 = 4.01 *10^-4


HA ---> H+ + A- and Ka = [H+][A-]/[HA] => 6.4*10^-5 = X^2/0.00802


=> x = [H+] = 7.16 * 10^-4 => mole of H+ = 7.16*10^-4 * 0.05 = 3.58 *10^-5



Now at 1/4 equivalence point 10^-4 mole of NaOH is added, => 10^-4 mole of A- is formed and 10^-4 mole of HA is consumed


=> mole of HA remained = 4.01-1 = 3 * 10^-4 mole


now [HA] = mole/total volume = 3 * 10^-4/0.051955 = 0.00577


[A-] = 10^-4/0.051955 = 0.001924


=> Ka = [H+][A-]/[HA] => 6.4*10^-5 = [H+] * 0.001924/0.00577 => [H+] = 1.92* 10^-4


hence pH = -log[H+] = 3.716


similarly u can find for 1/2 and 3/4 equivalence point