Question

9.You have 250 mL af a 0.425 M HNO, solution and you want to dilute it to exactly 0.200 M.How much water should add? you Mi 150mL Vi 0.175 250.0425 mf 0.200 Mr- 531.25 -2s0 ml A Mc 0. 200 O.200 Ne-0.200 2- 10) What volume of a 0.300 MH,SO4 solution is needed to completely neutralize 20.0 mL of a 0.200 MNaOH solution? Titation -3 2

Answer 1

If we require Vp mL of s, concentration of the stock solution to prepare Vq mi of Sq concentration. . solution, then vi, S., V₂, 5₂ are related by the following dilution formula. ... Vis, a Vg Sj a to oh.

9).'250 ml of 0.425(M) HNOZ solution is supplied... We have to dilute it to exactly 0.200 (M) ring solution Heres -250 ml. Vga? 5,=0.425(M) 5220.200(M). ve is the total volume of 0.200(M) Hino; solutions Vis, = V2S 2 » Va? Visio V2= 250 ml x 0.425 (x)... = 531.25 mL. 0.2004 We have to add (531.25-250) mL = 281.25 mL :: water to the 250 mL of 0.425 M) HNO 3 Solution this we get 1531.25 ml of 0.200 (m) HNO3 Solution. 10) H₂SO4 is diprotie acid. It dissociates 1 mole of thson. dissociated into a moles of htions in aqlleous solution. H₂SO,(as) ? 2H*(N2) + 50% (09). This 0.3 (M) H₂SO4 solution contains (2x 0.3) (M) :: 2 0.6 (M) Hitions Where as Naot is mono-basie bersic... 1 mole Nach dissociated into smole of onions in Solutions. Na ohbaz Nat (al) t or far).

Therefore, .0.2 (M) Naor solution contains 0.2 (M) of .. oh ions: intitration of strong-acid- Strong base, Htions. (mol) are neutralized by or iona ( moi) and vice versa. The net ionie rear of titration reaction (each) . V Ht(ag) + oH 61) H₂O(l) : so, volume of tttions (i.e H₂S og soln of required. .= ..20 m2 x 0.26417 (as per vis, =W252 for titration rean) 8, 3. Ò. 6(M) Htions coneentration = 0.3 (M) H, 50, Soln Vqz volume quired for titration Sta. BY 0-2 (M) Onion concentration'0.2(m) Nach soln. via volume of Nadh soln 2.20mL. 6.67 mL of 0.3(M) H₂SO4 is required to neutralized 20m2 of 0.2 (m) of NaOH solution: 0.6 CM = 6.67 m2