Question

Calculate the hypothetical pH AT THE EQUIVALENCE POINT for the titration of 50.00 mL of 0.0800 M Formic Acid (HCOOH, Ka=1.80x104)

Answer 1

At equivalence point com Hcoon + NaOH at H coo Nat +H₂O . moles of nwon = moles of Naon = mole of Hoo Na reacted Produced. moles of Hcoon = N( l) x Molarity = 50x 0.08 = 4410 mole 1ooo moles of or HCOONa = 4X10² Do Na is salt of weak acid and 80 this salt undergoes in hydrolysis pH of salt hydrolysis is given as strong Base for salt of weak acid & strong salt) where 2kw = lonisation cont L inks of water at 25°c Ka=dissociation const for aid. c = Cono of acid

- we have to calculate Conce of and, because after adding Nadh solution, volume of solution change to acid conc. also changes Hoor & Naon - VIMI=V2M₂ 5080.08 = V2X0.1 V₂=40 me .. total volume of solution = 50 ml of acid + 40ml of Bare 90ml e final faced conch] = 5610.00 98 i p = { [pikwapka_lopc] = } [14 -(-3.744) -log (0.04)] pha [ 17.744 - ( togoot) = 1 x 19.142 = 9.571 And