anewmanx
Harmless
Posts: 2
Registered: 8-5-2019
Member Is Offline
|
|
Need Phosphate Buffered Saline
I am having an incredibly hard time finding sterile phosphate buffered saline. It was missing from the at home lab kit for my microbiology class and
the company isn’t answering their phones. I take it the chemical is in short supply due to COVID testing.
By chance does anyone in the United States have 100-200ml available I could buy? I can pay with verified PayPal.
|
|
|
Harristotle
Hazard to Others
Posts: 138
Registered: 30-10-2011
Location: Tinkerville
Member Is Offline
Mood: I tink therefore I am
|
|
If you want to make, you need to achieve the following:
[PO4 -oids] = 11.8mmol/L
[Na+] = 157 mmol/L
[K+ ]= 4.5 mmol
I would just add
0.872g NaOH (21.8 mmol) (Draino, Bunnings. Get pure NaOH)
7.90g NaCl (135.2 mmol) (non-iodised salt from supermarket)
0.335g KCl (4.5 mmol) (no-sodium salt)
1.652g of 70%H3PO4 (11.8 mmol) (Rainex rust converter, Bunnings)
Make up to 500ml with distilled water
Adjust pH with an aquarium pH probe (what I use, it is cheap) to pH 7.4 using a dropper of NaOH (say 1g per 50ml-ish, you add drops of this so it
varies according to your situation).
Make up to 1l, then boil the snot out of it for 20 mins in the pressure cooker.
Cheers,
H.
Addendum: calcs
Table from Wikipedia for PBS 1x
Salt Concentration (mmol/L) Concentration (g/L)
NaCl 137 8.0
KCl 2.7 0.2
Na2HPO4 10 1.44
KH2PO4 1.8 0.24
Phosphate needed. 10mM Na2HPO4 + 1.8mM NaH2PO4
Source: hardware rust converter, 70% H3PO4
so need 0.0118mol x 97.994g/mol H3PO4 = 1.156329 g.
but H3PO4 is sold as a 70% liquid, which we must weigh, so
need 1.15632g x 100%/70% = 1.652g of hardware reagent.
NaOH needed (to convert H3PO4 to Na2HPO4 and NaH2PO4):
H3PO4 + 2NaOH --> Na2HPO4 + 2H2O (so need 2 x 10 mmol of that)
and H3PO4 + NaOH --> NaH2PO4 so I need 1.8mmol for that,
giving me a total NaOH requirement of 21.8 mmol.
mNaOH = n x mr = 0.0218 x 39.998 = 0.87195g ( but note that NaOH is deliquescent, so you will get less than you measure hence the need to adjust with
a dropper and a pH meter)
We need a total of 157 mmol Na, so we subtract the Na from the NaOH (157-21.8) and this gives us 135.2 mmol that we need to make up from NaCl. So
135.2 mmol = 0.1352 x 58.44 =7.90g NaCl.
Finally, we need 4.5mmol KCl
0.0045x 74.55 =0.335g KCl
|
|
|
anewmanx
Harmless
Posts: 2
Registered: 8-5-2019
Member Is Offline
|
|
Much appreciated! I got an email today saying it is in the mail, but having a backup is encouraging.
Thank you!
[Edited on 15-7-2020 by anewmanx]
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
