Suppose you have to prepare 1.00 L of the solution ( KH2PO4 and Na2HPO4 pH=7.31) and that this solution must be isotonic with blood (have the same osmotic pressure as blood). What mass of KH2PO4 would you use?
To determine the mass of KH2PO4 needed to prepare the isotonic solution, we need to understand a few concepts and steps involved in the calculation.
Step 1: Calculate the molar concentration of the isotonic solution.
For an isotonic solution, we need to match the osmotic pressure of blood, which is around 0.9% w/v NaCl (saline solution). Therefore, we can assume the concentration of our isotonic solution should be around 0.9% w/v.
Step 2: Convert the desired percentage concentration to molar concentration.
To convert the percentage concentration to molar concentration, we need to know the molar mass of the solute. The molar mass of KH2PO4 is:
[K(39.10) + H(1.01) + 2P(31.00) + 4O(16.00)] g/mol = 136.09 g/mol
Step 3: Calculate the molar concentration of KH2PO4.
Since the molar concentration is expressed as the number of moles per liter of solution, and we want to prepare 1.00 L of solution, we can set up the equation as follows:
Concentration (mol/L) = Mass (g) / Molar mass (g/mol)
We want the concentration of KH2PO4 to be 0.9% w/v (or 0.9 g/100 mL), which is the same as 0.009 g/mL.
Now, let's calculate the molar concentration of KH2PO4:
Concentration (mol/L) = 0.009 g/mL / 136.09 g/mol
Concentration (mol/L) = 6.61 x 10^(-5) mol/L
Step 4: Calculate the mass of KH2PO4 required.
We want to prepare 1.00 L of the isotonic solution. Multiply the desired molar concentration by the volume (1.00 L) to calculate the moles of KH2PO4 required:
Moles = Concentration (mol/L) x Volume (L)
Moles = 6.61 x 10^(-5) mol/L x 1.00 L
Moles = 6.61 x 10^(-5) mol
Finally, multiply the moles by the molar mass to obtain the mass of KH2PO4 required:
Mass (g) = Moles x Molar mass
Mass (g) = 6.61 x 10^(-5) mol x 136.09 g/mol
Mass (g) ≈ 0.009 g
So, you would need approximately 0.009 grams of KH2PO4 to prepare 1.00 L of an isotonic solution with a pH of 7.31.