you will need to prepare 100 mL of 0.025 M CH3CO2H (acetic acid), starting from

Question

you will need to prepare 100 mL of 0.025 M CH3CO2H (acetic acid), starting from commercial “glacial” acetic acid, which has a concentration of 17.4 M. Devise a method using a two-step serial dilution to make 100 mL of 0.025 M acetic acid solution. The glassware available is a 100 mL volumetric flask, 50.00 mL volumetric flask, 1.00 mL volumetric pipet, and a 10 mL graduated pipet. For the first step, dilute 1.00 mL of glacial acetic acid to 50.00 mL to produce “Solution 1”. You will then use Solution 1 to make the 0.025 M concentration for step 2. Determine the volume of Solution 1 needed to make 100 mL of 0.025 M acetic acid.

i got 870 M for solution 1 and 0.0028 mL but that doesnt seem right please help!

Explanation / Answer

This is serial dilution

in order to know unknown volume or concentration we use the following formula

M1V1=M2V2

concentration and volume before and after dilution

step 1

we have 1ml of 17.4 M glacial acetic acid and we have to prepare 50 ml of X concentration ,

so take 1 ml of 17.4 M glacial acetic acid into 50 ml volumetric flask and the dilute it with water , the concentration of final solution is calculated

M1V1 = M2V2

1*17.4 = M2* 50

M2 = 0.348 M (solution 1)

THEN step 2

we have to prepare 100 ml of 0.025 M solution from solution 1 (0.348M) , here we have to find the volume of solution 1 needed

M1V1 = M2V2

0.348 * V1 =0.025*100

V1 = 7.18 ml

take 7.18 ml of solution 1 (0.348M) in 10 ml graduated pipet transferred into 100 ml volumetric flask then dilute it with water

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