Question

Question 1 Abercan be prepared by two w Det each of the following Min 100 0 1 MH with 1000 05 MISS Ming 10001 MNL 1000 LIMNIU M O MON W 1000 LOS MNOH M1000 M LIMO Question 2 1 pts A 1.0L solution initially contains 0.3 MCH2COOH. Approximate the number of moles of CH3COOH and CH COO in the solution after 0.1 mol of solid NaOH is added, and calculate the pH of the resulting solution. K of CH3COOH is 1.8 x 105 1) The number of moles of CH.COOH present is Select . mol. Select 2) The number of moles of CH,COO presentis 02 3) The pH of the resulting solution is Select Question 2 1 pts A 10 L solution initially contains 0.3 MCH,COOH. Approximate the number of moles of CH,COOH and CH.COO in the solution after 0.1 mol of solid NaOH is added, and calculate the pH of the resulting solution. K of CH,COOH is 1.8 x 10 1) The number of moles of CHCOOH present is Select mol. mol 2) The number of moles of CH.COO present is Select Select 3) The pH of the resulting solution is select 01 Question 2 1 pts A 10L solution initially contains 0.3 M CH3COOH. Approximate the number of moles of CH3COOH and CH.COO in the solution after 0.1 mol of solid NaOH is added, and calculate the pH of the resulting solution. K of CH,COOH is 18 x 10 1) The number of moles of CHCOOH present is Select 1 - mol. 2) The number of moles of CH.COO present is Select 3) The pH of the resulting solution is select . 505

Answer 1

sont-Quentim 2: @ Buffer solutions - The compound which resist to change its pH by the addition of small amount of strong acid con streng base is called Buffer. 12 Yes, it is result'm buffer solm. Since He is a weak acid ģ RF is a strong base. 27 NH3 + NH4 Bor - yes it will result in buffer solm NH3 :-) NH4OH - weak base NhaBr-strong acid HENT KOH HeN.- Sveak acid Kot strong base ... yes, it will result in buffer soln 4) Helt kel Hel= strong acid keli salt No, it will not result in buffer solm 5 5) some of 0.IM HUN + Toome olmKOH HENT KOH KON 100% 0 m 1001010 lommoles lommole lommoles 2. No, it will not result in buffer som

deres tim Given that Ilq 0.39 CH3COOH û present & olmd q NaOH is added, . CHE COOH + NaOH - CH3COONa + H2O Initial ILX 0.30 alndes → 0.3 mdes 0.1 moles olmoles oil moles change Final 0-3 -0.) 20. 2 mdes 0.1 moles from Henderson- Hasselbalen tom is (salt pH = pka + log (Heid] Tog 1,8x105 Given Ka=181105 =) pkas -log ka -- log 1,8* - 5-login8 25-0.255 = 4.745 ..[salt] ..pH = pika + log - log Trid] 4 745 Log Sanj = 4.745+ 10966.5) pH = 4.745-0.3010 pH = 4.444 D). The no- g moles of CH36004 portent is 0.2 molo 22 The no. of moles q CH3COO present is Oal mole 3) The pH q resultant solution is 4.44 ..