What weights of NaH2PO4 and NaHPO4 would be required to prepare a buffer solution of pH 7.45 that has an ionic strength of 0.1? pKa2=7.12 , mwt NaH2Po4=120g/mol NaHPO4=142 g/mol
using the henderson-hasselbalch equation i got that the ratio of the concentration of NaH2PO4 to NaHPO4 is 2.14 but after that i'm stuck
5 answers
read the textbook
I am going to assume that you did that correctly. and go from there.
***The problem I believe is that you didn't realize what ionic strength is equal to molarity.
Let NaHPO4=A- and let NaH2PO4=HA
2.14=A-/HA
and HA +A-=0.1
Solving for A-,
A-=0.1-HA
Substitute one equation into the other to cancel out variables.
2.14=0.1-HA/HA solving for HA,
3.14HA=0.1
HA=0.318 moles
0.1-0.318 moles =moles of A-
Use molecular weights to solve for the mass needed for each one.
***The problem I believe is that you didn't realize what ionic strength is equal to molarity.
Let NaHPO4=A- and let NaH2PO4=HA
2.14=A-/HA
and HA +A-=0.1
Solving for A-,
A-=0.1-HA
Substitute one equation into the other to cancel out variables.
2.14=0.1-HA/HA solving for HA,
3.14HA=0.1
HA=0.318 moles
0.1-0.318 moles =moles of A-
Use molecular weights to solve for the mass needed for each one.
thanks!
Just remember that the masses that you solve for are only accurate for 1 liter.
no the answer above is wrong , u cannot assume[HA]+[A-]=0.1
recall back the formula of ionic strength ,then formed two simultaneous equation and solve it
recall back the formula of ionic strength ,then formed two simultaneous equation and solve it
3 answers