Thank you Philip for the kind comment. We could not have achieved this milestone without your active participation in the pharmaceutical calculations community. So, we appreciate your support as well.
@@PharmaceuticalCalculationsEasy Glad to be a part of this fantastic channel, may I request you to do some tutorials concerning calculations involving IU or MU for Drugs like Insulin , X-Pen in the future!
Hi, could you please explain how to go about the 2 questions below? -> How much Sodium Fluoride (NaF, MW = 42, F-, AW= 19) is required to prepare 200 mL of a solution such that 30 mL of this solution diluted to 1 litre of drinking water, will produce a final ] . concentration of F- of 1 part per million. (Note: The drinking water already contains 0.5 ppm F- ) ->Calculate the amount of Sodium Nitrite required to prepare 1L of a concentrated solution such that diluting 1 in 20 will yield 0.10 & w/v nitrite ion.(Nano2 MW=69, Na AW=23) Thank you so much!
Thank you Bimba for the comment. These questions have been previously solved on the channel. You may see the solutions here: th-cam.com/video/sBqOMJPEjBE/w-d-xo.html.
Thank you for watching and for the comment. The log portion is log ((0.0437-0.015)/(0.0375+0.015)) which is equal to log (0.0287/0.0525) and breaks down to log 0.547. That is where the -0.26 is coming from. Hope this helps.
Thank you the praise for watching and for the comment. The molar ratio of sodium acetate to acetic acid in the buffer solution is 1:1. This implies that you have one part of sodium acetate to one part of acetic acid. Hence, the total parts will be 1 + 1 = 2. To determine the moles of sodium acetate, you need to multiply the total number of moles by the mole fraction of sodium acetate. The mole fraction of sodium acetate is calculated by taking the parts of sodium acetate which is 1 and dividing it by the total parts which is 2. That is where the 1/2 is coming from. Hope this helps.
Congrats for 1 million views and keep the good content alive ...
Thank you Philip for the kind comment. We could not have achieved this milestone without your active participation in the pharmaceutical calculations community. So, we appreciate your support as well.
@@PharmaceuticalCalculationsEasy Glad to be a part of this fantastic channel, may I request you to do some tutorials concerning calculations involving IU or MU for Drugs like Insulin , X-Pen in the future!
@@philipraphael3676 Thank you for the comment Philip. We have added your request to the list of videos to be made.
@@PharmaceuticalCalculationsEasy thanks...happy to hear that!
how to determine when to add or subtract with regards to ph change calculations?? please and thanks .
Hi, could you please explain how to go about the 2 questions below?
-> How much Sodium Fluoride (NaF, MW = 42, F-, AW= 19) is required to prepare 200 mL of a
solution such that 30 mL of this solution diluted to 1 litre of drinking water, will produce a final ]
. concentration of F- of 1 part per million.
(Note: The drinking water already contains 0.5 ppm F- )
->Calculate the amount of Sodium Nitrite required to prepare 1L of a concentrated solution such that diluting 1 in 20 will yield 0.10 & w/v nitrite ion.(Nano2 MW=69, Na AW=23)
Thank you so much!
Thank you Bimba for the comment. These questions have been previously solved on the channel. You may see the solutions here: th-cam.com/video/sBqOMJPEjBE/w-d-xo.html.
@@PharmaceuticalCalculationsEasy oh thank you SO SO MUCH!! I understood nothing before watching your explanation!!
I am confused, please how do you know when to subtract the top and when to add when we are doing pH change. Thank you
thank you o much for all your effort, very helpful!
Glad you found the video helpful Nancy. Thank you for being an active member of the community.
how did the log become negative value (-0.26) in last exercise last question please??? it doesn't add up
Thank you for watching and for the comment. The log portion is log ((0.0437-0.015)/(0.0375+0.015)) which is equal to log (0.0287/0.0525) and breaks down to log 0.547. That is where the -0.26 is coming from. Hope this helps.
Sir on question 4, how did that 1\2 came about?
Thank you the praise for watching and for the comment. The molar ratio of sodium acetate to acetic acid in the buffer solution is 1:1. This implies that you have one part of sodium acetate to one part of acetic acid. Hence, the total parts will be 1 + 1 = 2. To determine the moles of sodium acetate, you need to multiply the total number of moles by the mole fraction of sodium acetate. The mole fraction of sodium acetate is calculated by taking the parts of sodium acetate which is 1 and dividing it by the total parts which is 2. That is where the 1/2 is coming from. Hope this helps.
@@PharmaceuticalCalculationsEasy thank you very much sir! I have understood