How do i prepare 0.1 ml KH2PO4-KOH buffer, PH 7.4 (20mM)?
Deoxyribose (2.8mM) - EDTA (0.1 mM) - H2O2 (1mM) - vaniline10 mM - phosphate buffer pH 6.9 - Phosphate buffer, 20mM - Phosphate buffer 0.1 M (pH 7.4) - Ascorbate (0.1mM). ( please i want the amount of acid and base by using pka)
To prepare a 0.1 mL KH2PO4-KOH buffer with a pH of 7.4 (20 mM), you can use the Henderson-Hasselbalch equation and the pKa values of the acid and base to calculate the required amounts. Here's the step-by-step process:
1. Determine the pKa values:
- KH2PO4: pKa1 = 2.12, pKa2 = 7.21
- KOH: pKa = 14 (since KOH is a strong base, pKa is considered to be infinity)
2. Calculate the acid to base ratio:
Since you want a pH of 7.4, which is closer to the pKa2 of KH2PO4 (7.21), you should use the second dissociation step of the acid.
- pH = pKa2 + log10([A-]/[HA]) (Henderson-Hasselbalch equation)
- Rearrange the equation:
log10([A-]/[HA]) = pH - pKa2
[A-]/[HA] = 10^(pH - pKa2)
3. Calculate the amounts of acid and base:
- [A-]/[HA] = 10^(pH - pKa2)
- Let's say you want a total volume of 0.1 mL. You can choose the ratio by considering the concentration of the buffer components. For example, you can use a 1:1 ratio of [HA] and [A-] at 10 mM each.
- Convert millimoles to moles by dividing by 1000:
[A-] = [HA] = 10 mM / 1000 = 0.01 mol/L
- Calculate the amount of acid (KH2PO4):
[HA] = 0.01 mol/L * (0.1 mL / 1000 L/mL) = 1 x 10^(-6) mol
- Calculate the amount of base (KOH):
Since KOH is a strong base, the amount required can be obtained by subtracting the amount of acid used:
[A-] = 0.01 mol/L * (0.1 mL / 1000 L/mL) = 1 x 10^(-6) mol
4. Convert the moles of acid and base to grams, considering their molar masses:
- Molar mass of KH2PO4: 136.1 g/mol
- Molar mass of KOH: 56.11 g/mol
- Amount of acid (KH2PO4) in grams:
mass = moles * molar mass = 1 x 10^(-6) mol * 136.1 g/mol = 1.361 x 10^(-4) grams
- Amount of base (KOH) in grams:
mass = moles * molar mass = 1 x 10^(-6) mol * 56.11 g/mol = 5.611 x 10^(-5) grams
Therefore, to prepare a 0.1 mL KH2PO4-KOH buffer with a pH of 7.4 (20 mM), you would need approximately 1.361 x 10^(-4) grams of KH2PO4 (acid) and 5.611 x 10^(-5) grams of KOH (base).