I am having trouble with this and don\'t think I\'m doing it right please help.

Question

I am having trouble with this and don't think I'm doing it right please help.

Thank you

7.3 Procedure Set up the following table in your notebook in which you collect your results: Sample #1 Sample w2 Sample #3 HCI-Standard mi(Na, Co) (g) buret intial (mL) Leol 2.06 31.05 buret final (m) 38.06 64.06 sample #1 25 mL 15 Sample #2 25 mL 03.50 Sample #3 25 ml 4.20 NaOH-Standared V(HCI) (mL) buret intial (mL) 33.50 buret final (mL) Antacid V(HCI) (mL buret intial (mL) Sample #1 25 mL Sample #2 25 mL 4.05 buret final (mL)

Explanation / Answer

Standardization of HCl sol. with Na2CO3.

I.

Na2CO3 + 2HCl = H2CO3 + 2NaCl

106 g ……..2 mol

0.601g……..x

X = 0.01134 mol/val HCl and also equivalents (val) of Na2CO3.

Same calculation for 0.602g:     x = 0.01136 mol HCl and also equivalents of Na2CO3 (1 mol = 2 val).

II

The results of titrations:

Sample 1

0.01134 mol HCl = (21.05-2.05)mL x 10-3 L/mL x MHCl

MHCl = 0.5968 mol/L and also NHCl = 0.5968 val/L

Sample 2

0.01134 mol HCl = (38.05-21.05)mL x 10-3 L/mL x MHCl

MHCl = 0.6670 mol/L (same value as val/L)

Sample 3

0.01134 mol HCl = (54.05-38.05)mL x 10-3 L/mL x MHCl

MHCl = 0.7087 mol/L (same value as val/L)

III.

The repeateability was very poor!

(0.5968+0.6670+0.7087)/3 = 0.65 N

Standardization of NaOH sol.

NNaOH = 0.65 N x (23.5-0.15)mL / 25 mL

            = 0.61 N (sample 1)

NNaOH = 0.65 N x (46.6-23.50)mL / 25 mL

            = 0.60 N (sample 2)

NNaOH = 0.65 N x (37.4-14.20)mL / 25 mL

            = 0.60 N (sample 2)

Average value = 0.60 N

7.4.3 Antacid tablet

i.

25 mL HCL used to dissolve the sample and 6.05 and 8.00 mL NaOH sol. used to titrate the excess acid.

6.05 mL x 10-3 L/mL x 0.60 val/L = 0.0036val NaOH used in titration ( and HCl in excess also)

8.00 mL x 10-3 L/mL x 0.60 val/L = 0.0048val NaOH used in titration ( and HCl in excess also)

Average value= 0.0042 val

ii.

25 mL HCL used to dissolve the sample

25.00 mL x 10-3 L/mL x 0.65 val/L = 0.0165 val HCl

iii.

0.0165val - 0.0042 val = 0.0123 val

Verify your data and improve your titration performance (the common mistake is the EP detection).

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