. A microbiologist desires to study the effect of pH on the growth of E coli bac
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Question
. 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 (HsPO), potassium dibydrogen phosphate (KH,PO) potassium hydrogen phosphate (K HPO) and potassium phosphate (KsPO.) to prepare the buffers. K 7.1110 K 632 x10 MATERIALS AVAILABLE Staadardized solutions: 1.00 MH,PO, 1.00 MKH PO, 1.00 MK,HPOs, 1.00 MK,PO, 1.00 MHCI, 1.00 MNaOH Solids: K,PO, KH PO,, 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 +0.0001-g) The acidic buffer she decides to prepare will have a pH of 2.000. She will use the phosphoric acid (1.00 MHPO.) and the solid potassium dihydrogen phosphate (KH POA) to prepare 100.0 mL of pH 2.000 buffer. She knows that a greater buffer capacity is obtained from solutions with higher concentrations and that for most laboratory-scale applications, a molarity of 0.500 is sarisfactory. Therefore, she deaides to make the buffer 0.500 M in phospboric acid. Why did she choose these reagents? what is the volume of 1,00 MHpo, and the mass ofKH2PO4 solid to be used to prepare 100.0 mL of the pH 2.000 buffer? A. B. 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 KHho, and K , 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 C. In the third experiment, she buffers the medium at a basic pH of 12.000 what is the molar concentration of the K HPO, if the desired molar concentration of the PO ion is 0.500 M D. What would the pH and pH change (ApH) be if 1.0 mL of 1.00 MNaOH were added to 50.0 mL of deionized water (pH 7.00)? Now, what would the pH and pH change (ApH) be that result from the addition of 1.0 mL of 1.00 MNaOH to 50.0 mL of the pH 7.000 buffer prepared in part B above? What would the pH and pH change (ApH) be if 10.0mL (no.0!) of ] .00 M HCI were added to 50.0 mL of deionized water (pH 7.00)? Now, what would the pH and pH change (ApH) be that result from the addition of 10.0 ml of 1.00 MHC) to 50.0 mL of the pH 7.000 buffer prepared in part B above? E.Explanation / Answer
Preparation of buffer
A. H3PO4/KH2PO4 (acid/base)
pH = 2.0
pKa = 2.15
(H3PO4) + (KH2PO4) = 0.5 M x 0.1 L = 0.05 moles
using hendersen-hasselbalck equation,
pH = pKa + log(base/acid)
2.0 = 2.15 + log(KH2PO4/H3PO4)
(KH2PO4) = 0.71(H3PO4)
or,
(H3PO4) + 0.71(H3PO4) = 0.05 moles
(H3PO4) moles = 0.05/1.71 = 0.03 moles
Volume 1.0 M H3PO4 = 0.03 moles x 1000/1.0 = 30 ml
moles KH2PO4 = 0.05 - 0.03 = 0.02 moles
mass KH2PO4 = 0.02 moles x 136.086 g/mol = 2.72 g
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B. KH2PO4/K2HPO4 (acid/base)
pH = 7.0
pKa = 7.2
(KH2PO4) + (K2HPO4) = 0.5 M x 0.1 L = 0.05 moles
using hendersen-hasselbalck equation,
pH = pKa + log(base/acid)
7.0 = 7.2 + log(K2HPO4/KH2PO4)
(K2HPO4) = 0.63(KH2PO4)
or,
(KH2PO4) + 0.63(KH2PO4) = 0.05 moles
(KH2PO4) moles = 0.05/1.63 = 0.03 moles
Volume 1.0 M KH2PO4 = x 1000/1.0 = 30 ml
moles K2HPO4 = 0.05 - 0.03 = 0.02 moles
Volume 1.0 M K2HPO4 = 0.02 moles x 1000/1.0 = 20 ml
dilute to 100 ml.
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