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
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.
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
To prepare a buffer solution with a desired pH and ionic strength, we need to calculate the amounts (weights) of NaH2PO4 and NaHPO4 needed. Here's how you can continue:
1. Calculate the concentration ratio of NaH2PO4 to NaHPO4 using the Henderson-Hasselbalch equation:
pH = pKa + log10([A-]/[HA])
Given pH = 7.45 and pKa2 = 7.12:
7.45 = 7.12 + log10([NaHPO4]/[NaH2PO4])
log10([NaHPO4]/[NaH2PO4]) = 7.45 - 7.12
log10([NaHPO4]/[NaH2PO4]) = 0.33
Taking anti-log (using base 10) on both sides:
[NaHPO4]/[NaH2PO4] = 10^0.33
[NaHPO4]/[NaH2PO4] ≈ 2.144
2. Now, we need to calculate the ionic strength (I) of the buffer solution. Ionic strength is a measure of the concentration of ions in a solution and is calculated as follows:
I = (1/2) * (C1 * z1^2 + C2 * z2^2 + ...)
Given I = 0.1, and since NaH2PO4 dissociates into two ions and NaHPO4 dissociates into three ions in solution, the ionic strength can be calculated as follows:
0.1 = (1/2) * (2 * C(NaH2PO4) * 2^2 + 3 * C(NaHPO4) * 1^2)
0.1 = (2/2) * (4 * C(NaH2PO4) + 3 * C(NaHPO4))
0.1 = 2 * C(NaH2PO4) + 3 * C(NaHPO4)
3. We can use the concentration ratio previously calculated to express the concentration of NaH2PO4 in terms of NaHPO4 concentration:
[NaH2PO4] = 2.144 * [NaHPO4]
4. Substituting this into the ionic strength equation:
0.1 = 2 * (2.144 * [NaHPO4]) + 3 * [NaHPO4]
0.1 = 4.288 * [NaHPO4] + 3 * [NaHPO4]
0.1 = 7.288 * [NaHPO4]
5. Rearrange to solve for [NaHPO4]:
[NaHPO4] = 0.1 / 7.288
6. Substitute the calculated [NaHPO4] value into the concentration ratio:
[NaH2PO4] = 2.144 * [NaHPO4]
7. Multiply the concentrations by their respective molecular weights to find the weights needed:
Weight of NaHPO4 = [NaHPO4] * mwt(NaHPO4)
Weight of NaH2PO4 = [NaH2PO4] * mwt(NaH2PO4)
Now, use the values obtained to calculate the weights of NaH2PO4 and NaHPO4 required to prepare the buffer solution.