Question

optional bonus problem) What volume of 3.25 M HNO3 is needed to neutralize 75.0 mL of 1.50 M Mg(OH)2

Answer 1

. The problem given is acid, base neutralization Seaction. thie law on neutralization formula es ativo = Na Vg — O no molmality f acid of base - W - volume of acid of base Ne y normality of acid or bare - volume f acid of bare If any, belongs to aid, then Naha belongs to have and viceversa. In problem at > 2H20 m 03) m m902 - a mol a mol My = apolumality f #wog. (given 3.25342. V = volunne q nog reruired to neuteline best Wq a Nohnality of mg (OH)₂ v2 = volume I may 04), Xonsumed in neutralization, in N₂ = MY Z, Ny Nohualitity of myott M + molasity ? my )y = 15m 27 v aloney 1 regte son of artumber on nuowed Neuhalization. (2, 2) N2 = M2 x 22 No = 1.592 W, → nemality ? ANO, Csequired No = M&Z, M.-) molanty of nog NE 23.25m x1 z, valaney f Htion of N; = 3.25 number of tion involved in neutralization Ne=3N

folum O ® . Now, aNzV2 . Substitute all values of Niitor, We and up then ^, = 3.25 N . Ng = 3N V2 = 75ml 13,25 xv, = 3473. Vi = 69.23 ml 3.25 M Anor required to of neutralize 69.23 ml egime of pism maglent),