Question

od luck! Question 15 4 pts A 20.0 mL sample of 0.115M H2SO3 is titrated with 0.1014 M KOH. At what volume (of added base) does the second equivalence point occur at? 50.0 mL 40.0 mL 22.7 ml 45.4 mL 20.0 mL Previous Next

Answer 1

For second equivalence point , the reaction is: H₂SO3 + 2KOH - K2502 + 2 H₂O- 00115 M H2S03 means - 1000 mL contains 00115 mole H₂SO2 20 ML 0.115 ( 1000 420) mole H₂SO₃- = 0.0023 mole H2SO₃. from the reaction stoichiometry, we can say that 0.0023 mole H2SO3 will react with 2x0,0023) = 0.0046 mole KOH,

so we need 0.0046 mole KOH. nowg 0.1014m KOH soin means - 0.1014 mole koH is in 1000 ml soln 0.0046 mole Koh will be in ( 1000 x0.0046) mL -0.1014 5 45.4 mb Hence, second equivalent point occurs at 45.4mL koH. Hences option -D is the answer.