how to prepare 6M sulfuric acid and 4M potassium hydroxide

Dear student
Please find the solution to the asked query:
Since you have not mentioned the volume of the solution to be prepared, we assume that we are preparing 250mL of each solution.

1) To prepare 6M H2SO4 solution 
The sulphuric acid present in the laborotary is normally 12 M,{you can check the molarity from the sulphurc acid bottle, sometimes it may be 16M also] so by dilution we can prepare the required molarity of H2SO4
​ and then using the formula:
If V1 = volume of concentrated acid
    C1 = concentration of concentrated acid
V2 = required volume of acid
C2 = concentration of acid which is required to be prepared.
V1C1 = V2C2
V1.12 = 250.6
V1 = 125 mL

So 125 mL of the concentrated acid is taken and 125mL of water is added to it to make the volume 250mL.

2) To prepare 4M KOH in 250 mL water
The weight of KOH required = molarity of solution required x volume of solution required x molar mass of KOH/1000
                                             = 4×250×561000
                                             = 56g

So when 56 g of KOH is dissolved in 250mL water, a 4M solution of KOH is obtained.

Hope this information will clear your doubts regarding the topic. If you have any other doubts please ask here on the forum and our experts will try to solve them as soon as possible.

Regards

 

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