Your automatic micropipette reads 006 from top to bottom. a. How many milliliter

Question

Your automatic micropipette reads 006 from top to bottom. a. How many milliliters will you pipette if you are using a P20? ____ b. How many microliters will you pipette if you are using a P1000? ____ c. What is the minimum amount that a P200 can accurately pipette? You are doing an assay in lab that requires you to prepare 100mL of 15mM NaOH (FW = 40) as your stock solution. How would you prepare this stock solution? Show your calculations: What is the pH of this solution? From the stock solution prepared in #2, you need to make a 100mL 3mM NaOH for a working solution. How will you prepare this solution? Show your calculations PNPP is a substrate for an enzyme that you are using in your experiment. You need to make 200mL of 4mM PNPP (FW = 371). How much PNPP do you need to weigh out? How you make the solution? Show your calculations

Explanation / Answer

1. A. P20 in a micropipette means that it can weigh till 20 microlitres (ul), and the recommended range for usage of this pipette is 2-20 ul. The reading 006 in this pipette equals 0.6 ul=0.6*10-3 ml= 6*10-4 ml. As such it is not recommended to pipette out 0.6 ul using a p20 micropipette, and a p10 micropipette would be ideal for that purpose.

1B. P1000 means that the micropipette can measure till a maximum of 1 ml (1000 ul) and the recommended range of usage of this pipette is 100ul-1000ul. the reading 006 in this pipette equals 60 ul= 60*10-3 ml= 6*10-2 ml.

1C. P200 means that the micropipette can measure till a maximum of 200 ul and the recommended range of using this pipette is 20 ul-200ul. So the minimum amount it can weigh accurately is 20 ul.

2. We have to make 100 ml of 15 mM NaOH solution. Now, this is given that the formula weight of NaOH is 40.

That means 40g in 1 litre=1 M= 1000 mM

Hence, 40 mg in 1 litre= 1 mM

Hence, 40 * 15 mg in 1 litre= 15 mM

Hence, 600 mg in 1 litre= 15 mM

But we only need 100 ml of this 15 mM solution.

so 60 mg in 100 ml= 15 mM

To prepare this solution, we would weigh 60 mg of NaOH, and dissolve it in water to make up the final volume to 100 ml.

3. pH of 100 ml 15 mM NaOH solution.

15 mM NaOH= 0.015 M NaOH solution which means that it would produce 0.015 mol of OH per litre of water

p(OH)= - (log 0.015)= 1.82 approximately

now, we know that p(H) + p(OH)= 14

Hence, pH= 12.18 approximately (by substituting the value of p(OH)= 1.82. we have to remember here that the volume of NaOH does not come in picture while calculating its pH.

4. To, prepare 100 ml 3 mM solution of NaOH from the 15 mM stock, we would use the formula,

M1V1=M2V2.

Now, Putting M1= 15 mM, M2= 3 mM, V2= 100 ml, we would get,

V1= (3*100)/15 ml =20 ml.

hence, we need 20 ml of the stock solution from question 2, and add 80 ml of water to make up for a 3mM NaOH 100 ml solution.

5. to make 200 ml of 4 mM PNPP (FW=371), We proceed like below:

371 g in 1000 ml= 1 M= 1000 mM

Hence, 371 mg in 1000 ml= 1 mM

Hence, 371*4 mg in 1000 ml= 4 mM

Hence, 1484 mg in 1000 ml= 4 mM

But we need only 200 ml of this 4 mM solution,

hence 1484/5 mg in 1000/5 ml= 4 mM

Therefore, 296.8 mg in 200 ml= 4 mM.

Hence, we would weigh out 296.8 mg of PNBB and dissolve it in water to make up the final volume to 200 ml.

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