When microorganisms reproduce, they release waste products that may change the p

Question

When microorganisms reproduce, they release waste products that may change the pH and prevent further reproduction. Therefore, culture media for growing microorganisms are usually buffered. The medium for the culture of lactobacilli includes a buffer prepared by dissolving 35.00g K_vd IPO_4, and 25.00h KH_2PO_4 (Ka = 6.2 times 10^-8; Kb = 1.6 times 106-7) in 250.0 mL water. Assume that potassium is a spectator ion in solution. What is the pH of this initial buffer solution? In one culture, it is found that the lactobacilli have added the equivalent of 0.016 moles of protons to the solution. What will be the resulting pH of the solution after buffering has occurred?

Explanation / Answer

buffer for

m = 35 g of K2HPO4

m = 25 g of KH2PO4

Ka1 = 6.2*10^-8

V = 250 mL

a)

the pH is given by henderson haselbach equation

pH = pKa + log(HPO4-2 / H2PO4-)

pKa = -log(Ka) = -log(6.2*10^-8) = 7.207

mol of HPO4-2 from

mol of K2HPO4 --> mass/MW = 35/174.2 = 0.200

mol of KH2PO4 --> mass/MW = 25/136.086 = 0.18370

so

pH = pKa + log(HPO4-2 / H2PO4-)

pH = 7.21+ log(0.200/ 0.18370)

pH = 7.2469

b)

H+ = 0.016 addition

recalculate pH

HPO4- + H+ = H2PO4

mol of HPO4-2 =0.200 -0.016 = 0.184

mol of H2PO4- =0.18370 + 0.016 = 0.1997

reaclaculate buffer pH

pH = pKa + log(HPO4-2 / H2PO4-)

pH = 7.21+ log(0.184/ 0.1997)

pH = 7.1744

decreases, since H+ are added

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