Standardization Of A Basic Solution DATA AND CALCULATIONS TABLE 1) HCI Concentra

Question

Standardization Of A Basic Solution DATA AND CALCULATIONS TABLE 1) HCI Concentration (M) 0.1115 Trial 1 Trial 2 Trial 3 2 Final Reading HC Buret (mL) ,g 3) | Initial Reading HCl Buret (rmL)"* | 0.00 = 0.00 = 0.00 4) Total HCI Volume (mL) 5) | Final Reading NaOH Buret (mL) 12212| 33 2| 49.82 6) Initial Reading NaOH Buret (mL)** 0.000.000.00 7) Total NaOH Volume (mL) 8) Volume Ratio HCl/NaOH 9) Molarity of NaOH solution (M) 10) Average Molarity of NaOH (M) 22 49-2 3-12 | 33,12 | 49,82 870 0928 .675 NOTE: You will need the average molarity of the NaOH solution for next week's lab. Record that value on the data table for Experiment 6 *x NOTE: The initial reading will only be 0.00 mL if this is what you initially start with. If you don't zero your buret initially, your initial reading will be whatever you first start with, for example, if you start with 0.05 mL, the initial buret reading for all three trials will be 0.05 mL for that buret.

Explanation / Answer

M1V1= M2V2

M1= molarity of HCl, V1= volume of HCl

M2= molarity of NaOH, V2= volume of NaOH

Trial 1 : 0.115 x 22.82 = 23.12 x M2

M2= 0.1135 M

Trial 2 : 0.115 x 32.22= 33.12 x M2

M2 = 0.1118 M

Trial 3 : 0.115 x 48.20= 49.82 x M2

M2 = 0.1112 M

Average molarity of NaOH =( 0.1112+0.¹118+0.1135)/3= 0.1121 M

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