1. A microbiologist desires to study the effect of pH on the growth of E. coli b
ID: 542331 • Letter: 1
Question
1. A microbiologist desires to study the effect of pH on the growth of E. coli bacteria. She plans to do the research by preparing culture media that are buffered at three distinctly different pH values. Therefore, she decides to use the triprotic acid phosphoric acid (H3PO), potassium dihydrogen phosphate (KH PO), potassium hydrogen phosphate (K,HPO) and potassium phosphate (KsPO,) to prepare the buffers K, = 7.11 x 10-3 Ko= 6.32 x 10-8 K,7.11 103 MATERIALS AVAILABLE Standardized solutions: 1.00 MHyPO,1.00 MKH PO, 1.00 MK,HPO, 1.00 MK,PO,1.00 MHCI, 1.00 MNaOH Solids: K,PO, KH POs, K HPO Volumetric Glassware: Pipets 1.00, 5.00,10.00, 25.00, 50.00 mL, Buret 50.00 mL, Volumetric flask 100.00 ml Analytical balance (weighs to ±0.0001g) A. of 2.000 er nat ior mo laboratory- to make the b PO, solid to be used to prepare For the second experiment, she buffers the medium at a neutral pH of 7.000 to minimize the effect of acid-producing fermentation. What volumes of the 1.00 M aqueous solutions of KH2PO, and K,HPO, must she take and dilute to 100.0 mL in order to prepare a pH 7.000 buffer that is 0.500 Min KH PO,? B.Explanation / Answer
for a pH = 7.0
we need to apply buffer equation
pH = pKa2 + log(HPO4-2 / H2PO4-)
7.00 = 7.21 + log(HPO4-2 / H2PO4-)
relate
HPO4-2 / H2PO4- --> K2HPO4 / KH2PO4
7.00 = 7.21 + log(K2HPO4 / KH2PO4)
10^(7-7.21) = (K2HPO4 / KH2PO4)
(K2HPO4 / KH2PO4) = 0.61659
K2HPO4 = 0.61659*KH2PO4
if we require
KH2PO4 = 0.5 M then
K2HPO4 = 0.61659*0.5
K2HPO4 = 0.308295
[K2HPO4 ] = 0.3083 M
[K2HPO4 ] = 0.5 M
if we need V = 100 mL = 0.1 L
mol of K2HPO4 = MV = 0.3083 *0.1 = 0.03083
mol of KH2PO4 = MV = 0.50*0.1 = 0.50
from molaritiyes given
V of K2HPO4 = mol/M = 0.03083 /(1) = 0.03083 L = 30.83 mL
V of KH2PO4 = mol/M = 0.50/(1) = 0.50L = 50.0 mL
then add V up to V = 100 mL
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