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Find average molarity of the NaOH solution and avg deviation We were unable to transcribe this...

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find average molarity of the NaOH solution and avg deviation
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on a separate sheet. tounding of) for all calculations for Part 3 on the bottom of this page or 3. Standardization of the NaO
We were unable to transcribe this image
on a separate sheet. tounding of) for all calculations for Part 3 on the bottom of this page or 3. Standardization of the NaOH Solution: Titration No. Acid buret, Final reading 1 12 0 24 48 Initial reading 26 24 12 Volume of acid used 120L izmi 0.0129 12mL 1 0.0129 1 12mL 0.0129 Moles of acid used 10.0129 Base buret, Final reading Initial reading 15,8mL sml 10. 8m asmL 32.4me 40mL is. 8m1 asme 32. Hori 9.2mL 7.4ml 18.60mL Volume of base used Moles of base used Molarity of base solution Average Molarity of the NaOH solution Average Deviation

Answers

You have used oxalic acid, which is a diprotic acid, to titrate NaOH, which is a monobase. The reaction between them is:

C204H22NaOH C20 2Nat 2H2O

Which means that per each mole of acid that has reacted, TWO moles of base reacted. This means that in the titration, the number of moles of NaOH in each trial is given by double the number of moles of oxalic acid; that is:

Titration 1, 2, 3 and 4: 0.0258 moles of NaOH

To calculate the molar concentration corresponding to each case, we must divide by the volume in liters, since the definition of molarity is moles per liter:

0.0258moles - 2.39M 1 M V(L) 0.0108L

0.0258moles - 2.80M 2 M V (L) 0.0092L

0.0258moles - 3.49M 3: M V(L) 0.0074L

0.0258moles - 3.00M 4 M V (L) 0.0086L

The average is given by the sum of all the values divided by the number of values (4):

2.39М + 2.80М + 3.49М + 3.00M М 2.92M 4

And the standard deviation is given by the sum of the squares of the differences between each value and the mean value, divided by N-1 and then all inside a square root. In this case, it would be:

1 [(2.39M 2.92M) (2.80M 2.92M)2 (3.49M 2.92M)2(3.00M 2.92M)2 OM = N- 1 0.46M

.

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