Question

13. Calculate the value for the equilibrium constant, Ksp for the following reaction. Ca(OH)2 (s) - > Cat? (aq) + 2OH(aq) assume that a 20.00 mL sample of saturated Ca(OH)2 was titrated with a 0.050 M standard HCI solution to the equivalence point. Assume that 8.1 mL of standard HCl was needed to get to the equivalence point

Answer 1

In the reaction Cacollis) - Cartag, +2012aq) In the Reaction, cacom), is titrated with her. So, During the titration at equivalent point No of mili equivalent no og meli equivalent of of You") ion Htion Cas ht neutralizes on ion) So, In the givers question Molarity of Hel M2 -0.0501 vol ay mal requindr = g.i me lets consider concor Molarity of QM-) ion= M, and as 20 ml oy Catoh), was Neutralized so vol oy (or-5 solution K = 8 20 ml j so from the ego no oy mili equivalent No oy meli equivalent y Htion(mmol) oy (oh) ion ( mmol) M,V, = Marc So, M,x 20 = 0.050X8.1 M. = 0.05018.9 = 0.02025 M 20 so to com os (04) ion cacOH)2 solution en is 6.029 25 m equilibrium constant (Ksp) at Simelary, In the Reaction the equilibrium is reporsented by " Ksp = [cat Icon1 [cacon ) as ca com) is salt so [cator), 2 = 1

Molente at equilibriun after the dissociation of ca6m). 2T (*] = fon'] so, equo [cat] = ron] . Keeping the talne of [catt) and cacorza en 90 Ksp = [(at2) rom-12 = TOH) [01-12 [Ca com)_] - Ž [or-73 as from the calculation conc y Tone) = 0.0 2025 | Kcp (OOdoas 3 = 4.152 x 10-6 Reaction will be 4.152x10-6 1. Ans, Ksp for the following