Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

I am not sure how to go about #1 Is #2 correct? Not sure about my work in #3 I t

ID: 477415 • Letter: I

Question

I am not sure how to go about #1 Is #2 correct? Not sure about my work in #3 I tried to work out the first part of #4, but I'm not sure if it's right or how to answer "how I would make this dilution." All help appreciated! 5. Dilutions: A. Your lab has a stock solution of 12 NHCI acid (MW 36.5 g/mol). However your experiment requires 100 ul of 3 NHCI acid. How would you make this dilution? B. You added 300 mL of diH20 to a 50 mL solution of Nazso4 with an initial concentration of 0.3 M. What is the final concentration of Na2SO4 in mM? 2 3Sou. C. How would you make a 320 mL of a 15% ethanol from a 80% ethanol? uso added to 330 mL water 20 320 D. You prepared a stock MES salt (MW 217.2 g/mol) buffer by adding 30 g of MES salt to 250 mL of diH20. For your experiment, you need 300 mL of a 50 mM MES salt buffer. What was the initial concentration on the stock MES buffer (in M)? How would you make this dilution?

Explanation / Answer

On dilution, we add only solvent , but the amount of solute remains the same.

Thus on dilution, the millimoles of solute before dilution = millimoles of solute after dilution

which is V x M before = V x M after dilution.

a) You have stock solution of 12 M HCl solution. You want 100 mircrolitres of 3 M HCl.

We need to calculate how much of stock solution need to be diluted to make 100 microlitres of 3M .

Thus V x 12 M = 100 x 10-6 L x 3 M

Hence V = 25 x10-6 L = 25 micro litres

Take 25 microlitres of 12 M HCl and dilute with 75 microlitres to get 100 microlitres of 3 M HCl.

b) You have 50mL of 0.3 M solution.

Added 300 mL of water, so that the total volume now is 350mL. the molarity of this solution is to be calculated.

50mL x 0.3 M = 350mL x M

Thus molarity of diluted solution = 0.04285 M

c) you have 80% ethanol which means 80g of ethanol present in 100 mL of solution.

the molarity of 80% ethanol = moles of ethanol /V (L)

= (80/46) /0.1L

= 17.39 M

Now we need to dilute it to 15% ethanol

molarity of 15% ethanol = (15/46) /0.1L

= 3.26 M

Volume of 15% ethanol required = 320 mL

Thus using VM before = VM after dilution

V x 17.39 M = 320mL x 3.26 M

thus V = 59.98mL

d) molarity of stock buffer = (30/217.2) /0.250L

= 0.5525M

you need 300mL of 50mM (= 50x10-3 M) solution

V x 0.5525 M = 300mL x 50 x10-3 M

V = 27.15mL

Take 27.15 mL of stock solutio n of molarity 0.5525M and silute it to total volume of 300mL to get the 50mM solution.