Question

150 mL of 10-1-(WA) SDS 1w/veg of solute/100 ml 10g: 100 ml 150 ml 10g) 15g SDS in 150 ml dd Haa (100ml) b. 200 mL of 250mm tris buffer, por 7.5 Tris 1210 g/mol PH 7.5 es pot - 14-7.5 = 6.5 Pka of tris: 205 -> pkb - 14-pka = 5.94 POH pkb + log [ BH"] Balt [B]-* [et] Tel : antilog (6.5-5.94) - 3.631 [4] . 3.631 [8] [BHT] + [b], 250 3.631[B]+[B] -25'0 4. [e]. 53.986 MM 250M-53.99 M 1960 MM [BH]: 3.631 [6]. 196.025 MM 250 mM 200 ml tris Base molarity Mass molarmass x volume maaso x10 61 6.0559 quo ml of 196.025 mM HCI Mi Vi - Mova 1M XV, 196.025x10 3 Mx 200 mL. Vuel 39.205 mL Buffer: 6.055g This : 39.205 mL IN HCL 200mL +'39,205 = 160.795 mL de Hoo

can someone help with #2

Answer 1

Answer a) 150 ml of 10% (W/V) SDS 10 % (W/W) SDS => 100 ml Solution contains log SDS : 150 ml solution contains 100x1509 => 159 Sos Msos=159 diluted dissolved in some of Water b) 200 ml of 250 mM Tris Buffer pH = 7:5 => POH= 14-PH = 6.5 pka of tris =8.06 1 pky = 14-pka = 5.94 POH = pkb +log[1347) >Dalt [B] y Base [BH'] antilog (6.5 - 5.94) [13] =3.631 [BH+] = 3 631 [] -A150. (BH1]+[B] = 250 3.631[6]+[3] = 250 [B] - 53.986 MM 1 [BH] = 3.631 [6]=196.095 mm

Need : Tris 애 HO POH | HOT oH + HCl + HO VIVOH 06 - NHỰA NHỎ (n) (BH*) 1-0: 250 196.025 eqn: (250–196.0251 196.025 MM 53.975 MM

Need: 250mm zoome It's Base Mass Molavily molemassxvolume M : 250x1623 s *1212 not 00x1003 2 M=660559] 200ml of 196.025mm Heh MY Stock Hel M2Y detuted 14 xva 196.025 x 103 x200 M ML [rug = 39. 205m2) (for IN ALL, Molarity = nomality) 10=1M _ 6.055g Tois Base Buffer - 39.205ML IN HCL 200-39.205) ML 160.795 waler

© 3oome of looppm thy mine 100ppm => loongh => AL Solution contains loong thymine (1000mL) 300ml Solution contains to bot Boong =>(30 mg Chymine @ 400mL of BoomM Cache mash : 300 x 10 MOT x 147.02x400 x 103 met M= 47.0469] present . in 4oome of distilled water. @ Some of 1MM selente solution Nag Seoz-> 2Nut + seo? M = 4x0 B me x172.9 cm x 5ox 10°3 z M=0.0086459 Naz 503 dissolved in some water ③ 500ml of 100mm phosphate Buster pH = 70 H₂ Pou Pka=702

pH = pka a dog [mpoy? ] – salt 1 [m₂ pou J arid 70 - 7.2 + loСироч? 1 Lч, Роч). >Снео2-1 - 0-6 30*4 Сп, ғоч ] Ass [nPou-]+[и, ғоч ]- loo о• L304 «Сирғоч ]+ Сн, Ра] = loo Сп, Роч ] : &1. 314мм (Lпроц?-] - o-63096 СиPoы] : 36-6&LMм Na2HPO4 + HCl 2 Nahapout rach t-o: loo» 6-314 А 61: 314 мм 4* : Cipo - 6 -3ч) о 7 36 686 мм

Poy Need: 500 ml of 100 mM Naz Hpou mass - 100X10mol x 141969, 500X10 VOX10"} m=70989 500ml of 61.314 MM HCL Milli = M242 stock . diluted TMKU = 61-314 x103 of x 500mL [Istock = 30.66 mL of IN HC L Buffer -7.0989 NOZHPOU < 30.66ml of 1 NHCl "(500-30.66) => 469.34 ml. distilled Weiter