For the Last part of the problem I do not know where the 200mL for M1v1=M2V2 cam
ID: 1044704 • Letter: F
Question
For the Last part of the problem I do not know where the 200mL for M1v1=M2V2 came from. I get the rest of the problem except that part. Thank you in advice!
? csu F Portal Ho Home Chegg.c( Course: Spring 2 1 Inbox (583) Buffers, Titration Thermodynamic ? Buffers, Titra + v can file:///C/Users/carol/AppData/Local/Packages/Microsoft. MicrosoftEdge_8wekyb3d8bbwe/Tempst ? ??? ers, %20Titra tions, %20Th You have 100.0 mL of a 0.250 M NHs sclution that you are going to titrate with 0.500 M HCI solution The pkb of H,CCOOH is 4.75 What is the pH of the original NHs solution? 0.250 E 0.250-x 10."-10-475-1.78x 10-5 1.78x10 x=2.11x10-a Check poi--log12.1 1x10")-2.77 pH = 14.00-2.77 = 11.32 What is the pH at the half equivalence point AND what volume o the HCl solution is required? Al the haf-equivaence pont, we have added erough HCI to react with talf of the original ammonia 0.100 L NH.x 0.0250rnol HC-X 1L-=0.0250 L HCI-25.0 mnLHCI Al the half equivalence cont, the pHpk, 14.10 pk, 1400-4.75 9.25. What is the pH at the equivalence point AND what volume of the HCI solution is required? At the equivalence point, we have added enough HCI to react with al ot the onginal base 0.100 L NH,x 0.0250 mo! HClx 0.250-x 0.250 2.11x10 x 1 00 = 0.843% 0.250 mol = 0.0250 mo? NH, 0.500 mo! 0.250,rol = 0.0250 moll. 0.0500 L NCI:50.0 mL HCl 0.500 mo [NH:? 0.0250mol 0.150L =0.167 M In the last line, we remembered to take into account the fact that the total volume of sclution has changed due to the additionof acid ENG :37 PM Type here to search US 4/13/2018 2Explanation / Answer
Initially you are starting with 100 ml of NH3, so V1 = 100 ml
To this you are adding 100ml of HCl, so V2 = 100 ml
Total volume after mixing = V1 + V2 = 100 + 100 ml = 200 ml.
Out of the 100ml of HCl that you are adding, you have calculated that 50 ml is used for neutralising NH3. therefore excess HCl = 50 ml.