Question

Name CHM 1202 Titrations and Polyprotic Acids- Reading Guide Polyprotic Acids (section Polyprotic acids contain more than one ionizable per molecule. For all polyprotic acids, a ionization process occurs. Give the chemical equation for the first and second ionization of hydrosulfuric acid, H,S. First ionization: HS (aq) + H20 (1) Second ionization: In the above example, is the dominant producer of H,0*, and thus [H0') = Titration Curves (section Refer to Figure 14.23 to sketch the following titration curves. Label the equivalence point. 1) Strong acid to which strong base is being added 2) Weak acid to which strong base is being added

Answer 1

C # Polyprotte acid section • Polyprotic acid contain more than one imisable protons 14304) per molecule • For all proly protic acid a stepwise ionisation process occurs First imisation : H2S (ag] + H20 () ► stenst H28 (ae) (pka = 6-2) Second imisation : SH+ay+ Holly —> s?ice) + 3 % caj (pKa = 1294 . the above example, H2S is the dominant producer of 1309 and thus [H30] = [H25] Actually ka = [ 13025617 HEVAL [H25] c 11 (430*] Ka Chan # Titration curve : Og® 1 C o strong acid to which strong base o is added wears aud to which strong base is being added pH of 12 analyk solutim O weak aud Strong base -khoacd- Strong base 250 50 75 Volume of base added (m2)

* Weak base to which strong acid is being added Equivalance point volume of acid added → ☺ Poly protte acid to which strong base is being added. I do acord BH -11.9 V e 2nd Equivalance point pH=pkaz ya 2nd Halt equiralance point [H4] = [42"] 7 $H=61 te ist equivalance point x ist Half equirulance point: [itz 4] = [HA] TpH=pkar 2 Titration of My A by NaOH : Naoh +H2A —— NaHA + H2O (Kar) - 0 NOOH + NaHA → Na2A + H20 (Kazl - of Kai Y, Kazx109 then reaction o will complete before reaction @ starts, we will get two well separated equivalance point as shown in the graphi