Question

Part A As a technician in a large pharmaceutical research firm, you need to produce 300 mL of a potassium dihydrogen phosphate buffer solution of pH = 7.07. The pK, of H2PO4 is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO stock solution, 1.50 L of 1.00 M K HPO, 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) MÅ mo a? Volume of KH, PO needed = Value Submit Previous Answers

Answer 1

Part A : Volume of KH₂PO₄ needed = 174 mL

Part B : pH = 7.61

Explanation

Part A :

Let volume of KH₂PO₄ needed = x

then, volume of K₂HPO₄ = (total volume) - (volume of KH₂PO₄ needed)

volume of K₂HPO₄ = 300 mL - x

moles KH₂PO₄ needed = (volume of KH₂PO₄ needed) * (concentration of KH₂PO₄)

moles KH₂PO₄ needed = (x) * (1.00 M)

moles KH₂PO₄ needed = x mmol

Similarly, moles K₂HPO₄ needed = (300 - x) mmol

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [weak acid])

pH = pK_a + log(moles K₂HPO₄ needed / moles KH₂PO₄ needed)

7.07 = 7.21 + log((300 - x) / x)

log((300 - x) / x) = 7.07 - 7.21

log((300 - x) / x) = -0.14

(300 - x) / x = 10^-0.14

(300 - x) / x = 0.724

300 - x = 0.724x

300 = x + 0.724x

300 = 1.724x

x = (300) / (1.724)

x = 174 mL