Part A 300. mL of a potassium dihydrogen phosphate buffer As a technician in a l
ID: 1032252 • Letter: P
Question
Part A 300. mL of a potassium dihydrogen phosphate buffer As a technician in a large pharmaceutical research firm, you need to solution of pH -6.93. The pK, of H2 PO4 is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2 PO, stock solutión, 1.50 L of 1.00 M K, HPO4 stock solution, and a carboy of pure distilled H2O How much 1.00 M KH,PO, will you need to make this solution? (Assume additive volumes.) Express your answer to three significant digits with the appropriate units. > View Available Hint(s) Symbols Volume of KH2PO4 needed- Value || Units SubmitExplanation / Answer
a. As we know that the formula to calculate the pH of a buffer solution is given by:
pH=pKa+log([A]/[HA])
Here, HA represents the acid which in our case is H2PO4- . The pKa value of the given acid is 7.21. We also know that the pH of the solution is independent of the dilution. The pH of the resulting solution should be equal to 6.93 as given in the question. Putting the values in the equation, we will get
=> log([A-]/[HA]) = -0.28
=> [HPO42-]/[H2PO4- ] = 0.756
=> Since the concerntration of both the stck solutions are same, we will just have to worry about their volumes.
The ratio indicates us thatb for every XmL of H2PO4- we will need 0.756X mL of HPO42- . Now total volume is 1.756X wchich should be equal to 300mL because volume is additive. Therefore, X = 170.84mL.
b. In the equation given to us in the question, In the normal conditions,
pH = 7.4
pKa = 6.1
[HCO3-] = 24mM
Putting these values in the equation and solving we will get the partial pressure = 0.218 Atm = 165.68 mm Hg
Now, the question asks about the pH when the pressure equals 23 mm Hg. Again solving the equation with
pKa = 6.1
[HCO3-] = 24mM and
Partial pressure = 23 mm Hg,
We will get the value of pH = 7.524.
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