Question

ensune prper 4. What is the function of an indicator in a titration? 5. How many milliliters of 0.258 M NaOH are required to completely neutralize 2.00 g of acetic acid HC2H3O2? 6. What is the volume, in milliliters, of a 0.350 M KOH needed to completely neutralize 15.0 mL of a 0.250 M H2SO4 solution? 2KOH(aq)+H2SO4 (aq) K2SO4 (aq) + 2H,O() 231

Answer 1

The function of an Indication :- A chemical substance that changes color in response to a chemical change. During the titration a drop of an indicator is added at the beginning of titration. The indicator changes the colon of solution when the end point has been reached. 159) chemical balanced equation between NaOH + CH3COOH NgaOH a + CH3COOH (9) => c ,cac cap + i mol 4) mol i mol To reach the neutralization point I mol ole Naot required i mol of chacoolt. At the end point Number of moles No. of moles of of Naott (ni) - CH3COOH (na) molar mass of acetic acid = 60.052 g/mol mass of acetic acid = 2.00 a moss (9) moles ot CMzc001 molar mass (8/mol) Number of

CH3COOH moles : 2100(9) 60.052 (8) moll na nz : 0.03330 moles the 0103330 moles ot CH3COOH required, 0.03330 moles of NaOH to reach end point. No.et moles ob Nooh (11) Molarity (My a Naon Volume ob NaOH V solution () 0. 258 M · 0.03330 mod. Vi (L) Vi L2 - 9 0.03330 mol 0.129 Litre. 0. 258 M. 1 ml = 0.001 Letre 0.129 Litre : 129 mL so 129 mL of Naot required (6A) 2 KOH (99) + H₂SO Le cag → Kagogae + QHqoll a mol Imol

I mol of Hasou required 2 mol of RoH IM = "/vo] where M: molarity n= no ob moles V: Volume . Hasou y = 1500 ml (on 0.0150 L M, = 0.250M. ; n, .? [n, M, XVI] = 0.250m X 0.0150L = 0.00 375 mdes 0.00375 moles of Hasoy required 280.00375 moles . = 0.00750 moles of Kolt. 0.00750 mol i me e mire V 13 01 350 M = 0.0214 Litre (ora 214 mL -titration required 21.4 mL of koH.