Question

2) If 24.7 mL of 0.250 M NaOH solution are needed to neutralize 19.8 mL of H,SO, solution, what is the molarity of the H,SO 2Nadd tuSot Naso2H20 24 NOS 25.0 g of 5.0 % ( by mass ) acetic acid solution are titrated with 0.300 M NaOH. What volume of NaOH will be needed to neutralize this sample? 3)

Answer 1

NaOH - VNaOH = 24.7 ml [Mach] / Mnaoh = 0.250M 3 Molarity of Naoh WH, son = 19.8ml Muy son = ? 2 N2014 * H? Soy — Ne, soy + 01 At Equivalence porest, milliaq of H₂SO4 2 millicq of Nach point Neutralisation millimor xn-factor) k sou a mmoles ' Naokx in factor > 19.8ml x A NaOH sou x 2 = 24.7ml x 0.25X1 n-factor of hy sou = & [Musou = 0.156 M) ( as It furnishes 2 H * in solution So its basicity is 2) n-factor of Naoth is 1 as it pornishes only one oft in solutica 3 259 85% CH, Cool - 25 x 5 hog a given mass of Cheon - acetic acid Molar mass of gx127 4X1+2X16 acetic acid = 60 g/mole men may of CH COH = 1.25 -0.0208 moles no. of moles of CH₂ Loon = given mass of Ch Cooh Molar mass of CM, cooh . To

Nino, at equivalence paint , millies of Ut curt = millieq of Noor mmoles of Ch Ceolt X(n. Ja too ea crone = millies of Naoll (0.02.08 moles x1000) xi- mollege of Naoh 2018 = 073M X Yaore (me) x 1 [ 69.33 und = VNooh immoles