hello i have a question about my buffer an pH LAB SO i am suppose to find the in
ID: 497251 • Letter: H
Question
hello i have a question about my buffer an pH LAB
SO i am suppose to find the inital moles and ratio.
pH
.20M HC2H3O2 2.97 used 25mL of it and diluted with DI water 50mL
.20M NaC2H3O2 6.30 same
Combined two 4.85
combination is below
pH
trial vial #
1 4.84
2 4.47
3 4.90
4 3.20
5 11.67
6 4.4
7 5.43
8 5.14
9 4.32
10 4.70
11 5.20
SO i found the concentration using M1V1/V2= .20M.25mL/.50mL =.90M FOR BOTH
now i have to find the inital moles for each trail, do i us the new mL combination for each trail and how to i solve it. also how do i find the ratio?
10.Manual pdf Xe Masteringchemistry Ais xG how to take a screensho x C Chegg Study IGuided S G What is the con X 102 M ants-chem t/Classes/102/manual/102 Manua 5 41 131 102 Manual.pdf 12. Use clean 10 mL graduated cylinders and follow the chart to carefully measure the volume of each reagent indicated into separate, well-drained shell vials. Shell combined 20-M O.200-M o 10-M Vial sodium Number Solution acetic acid acetate 0.10-M NaOH 3.0 mL 2.0 mL 2.0 mL 3.0 mL 4 3.0 mL 3.0 mL 3.0 mL 2.0 mL 4.0 mL 2.0 mL 3.0 mL 3.0 mL 4.0 mL 2.0 mL 3.0 mL 1o 2.0 mL 4.0 mL 11 12 Cover each shell vial with parafilm Miv the contents well hv inversion then meas 91% 1:30 PMExplanation / Answer
Henderson-hasselbach equation can be used to calculate the ratio of moles of HC2H3O2 and its conjugate base ,NaC2H3O2 .
pH=pka +log [conjugate base]/[acid]
or, pH=pka +log [NaC2H3O2]/[HC2H3O2]
pka for acetic acid =4.75
trial 1 ,pH=4.84
pH=pka +log [NaC2H3O2]/[HC2H3O2]
or, 4.84=4.75 +log [NaC2H3O2]/[HC2H3O2]
[NaC2H3O2]/[HC2H3O2]=1.230 [molar ratio]
concentration using M1V1/V2= .20M* 25mL/50mL =0.1M
As both NaC2H3O2 and HC2H3O2 used had the same molarity,so, [NaC2H3O2]/[HC2H3O2]=1.230=0.1M*V1/0.1M*V2=V1/V2
So,V1/V2=1.23 or, V1=1.23*V2
if total volume =4.0 ml=V1+V2
then 4.0 ml=1.23V2+V2=2.23V2
V2=4.0/2.23=1.8 ml
V1=4ml-1.8ml=2.2 ml
So V1=volume of NaC2H3O2 =2.2ml
V2=volume of HC2H3O2=1.8ml
So initial mole of NaC2H3O2=0.1M*2.2ml=0.1 mol/L *2.2*10^-3L=2.2*10^-4 moles
initial moles of HC2H3O2=0.1M*1.8ml=0.1 mol/L *1.8*10^-3L=1.8*10^-4 moles
similarly other trials can be solved
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