How much Tris-HCl would you need (in grams) to add to create 100 mL of a 0.223 M Tris buffer that is pH 9.64. The pKa of Tris-HCl is 8.30.
I'm on my last try and thought I had it multiple times but keep getting it wrong.
Why not post what you've done and let us help find the error.
To calculate the amount of Tris-HCl needed to prepare the buffer, you need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH = desired pH (9.64 in this case)
pKa = ionization constant of Tris-HCl (8.30 in this case)
[A-] = concentration of the deprotonated form of Tris-HCl (Tris-base)
[HA] = concentration of the protonated form of Tris-HCl (Tris-HCl)
1. Calculate the ratio [A-]/[HA] using the Henderson-Hasselbalch equation:
9.64 = 8.30 + log([A-]/[HA])
Simplifying the equation gives:
1.34 = log([A-]/[HA])
2. Convert the ratio into a linear scale:
10^1.34 = [A-]/[HA]
Approximately:
25.1 = [A-]/[HA]
3. Since the total volume of the buffer needed is 100 mL and the concentration of the buffer should be 0.223 M, you can use the definition of molarity:
Molarity = moles of solute / volume of solution (in liters)
0.223 = (moles of Tris-HCl) / (0.100 L)
4. Assuming that the total volume of the buffer is the sum of Tris-HCl and Tris-base, the concentration of Tris-base ([A-]) can be calculated as:
[A-] = (25.1 / (25.1 + 1)) * 0.223 M
Simplifying:
[A-] = (25.1 / 26.1) * 0.223 M
[A-] = 0.214 M
5. Since the molecular weight of Tris-HCl is 157.60 g/mol, you can calculate the moles of Tris-HCl needed using its concentration:
moles of Tris-HCl = (0.214 M) * (0.100 L)
6. Finally, calculate the mass of Tris-HCl required using its molar mass:
mass of Tris-HCl = moles of Tris-HCl * molar mass
mass of Tris-HCl = (0.214 M) * (0.100 L) * (157.60 g/mol)
mass of Tris-HCl = 3.36 g
Therefore, you would need approximately 3.36 grams of Tris-HCl to prepare 100 mL of a 0.223 M Tris buffer at pH 9.64.