What is the pH at the equivalence point in the titration?

Calculate the pH at the equivalence point in the titration of 25.0 mL...?

• Calculate the pH at the equivalence point in the titration of 25.0 mL of 0.250 molL-1 acetic acid (CH3COOH) with 0.0500 molL-1 NaOH. For CH3COOH, Ka = 1.80 x 10-5. Any help would be appreciated. Preferably a guide through it XD

• Answer:

  at equivalence pt mmole of acid =mmoles of base mmole= mL x M sooo mmoles pf acid = 25 mL x 0.25M = 6.25 mmoles Vol base x M base = 6.25 mmoles vol base = 6.25 mmol/0.05 M = 125 mL soo total volume = 25mL + 125 mL =150 mL In the reaction CH3COOH + NaOH ----> CH3COO^-1 Na+ + H2O at EP all the mmoles of acid have been converted to an equal number of mmoles of sodium acetate, CH3COO^-1 Na+. So the mmoles of sodium acetate produced is 6.25 mmoles but now those mmoles are in 150 mL total volume.The molarity of the acetate is then 6.25 mmoles /150mL = 0.0417 M acetate is the conjugate base of a weak acid ( acetic Ka = 1.8 X 10^-5 ) which means that acetate is a pretty strong base and in water it will hydrolyze to produce small amounts of OH^-1. Since OH^-1 is a strong base, the pH will rise to ~ mid 8s. CH3COO^-1 + H2O ---> CH3COOH + OH^-1 The amount of OH^-1 and subsequent pOH , then pH can be determined from the hydrolysis constant which is Kw/Ka =Kh :: or 1X 10^-14/1.8 x 10^-5 = 5.56 X 10^-10 sooo of the 0.0417 M acetate a small amount ( X ) will hydrolyze to produce X amounts of CH3COOH and X amounts of OH^-1 soo using the Kh Kh = [x][x]/0.0417 -x x will be small compared to 0.0417M and can be neglected in the denominator...soo X^2 = 5.56 X 10^-10 x 4.17 X 10^-2 x^2 = 23.185 X 10^-12 ::: x= 4.815 X 10^-6::: pOH = -log [OH^-1]= - log (4.185 X 10^-6 )= 5.38 ,,, pH =14 - 5.38 = 8.62