Question

Cabrillo College Chemistry 1B Buffer Problem Set 1. A desires study the effect of pH n 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 phosphoric acid (H,PO), potassium dihydrogen hydrogen phosphate o") and potassiam phosphate POe) to prepare the buffers K-1 7.11 103 Ka- 6.32 x 10" Kal 7.11 x 10 MATERIALSAFALLABLE Standardized solutions: 1.00 MH3PO, 1.00 M KH2PO4 1.00 M K HPO, 1.00 MKAPo, 1.00 M HCl, 1.00 MM NaOH Solids: KBPO4, KH2POe, K2HPO4 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 20.0001-g A. The acidic buffer she decides to prepare will have a pH of 2.000. She will use the capacity is obtained from solutions with higher concentrations and that for most laboratory-scale applications, a molarity of 0.500 is satisfactory. Therefore, she decides to make the buffer 0.500 Min phosphoric acid. Why did she choose these reegents? What is the volume of 1.00 MH3PO4 and the mass ofKH2POu solid to be used to prepare 100.0 mL of the pH 2.000 buffer?

Answer 1

Molarity of buffer = 0.500 M Volume of buffer = 100.0 mL Molarity of H_3PO_4 = M PH = 2.000 Using Henderson-Hasselbalch equation, pH = pKa + log [Conjugated Base]/[Acid] 2.0 = 2.148 + log [Conjugate Base]/[Acid] [Conjugate Base] = 0.711[Acid] equation 1 [Acid] + [Conjugate Base] = 0.500M equation 2 solving equation 1 and 2, [Acid] = 0.292 M [Conjugate Base] = 0.208 M Using dilution equation, M_1 V_1 = M_2 V_2 V_1 = M_2 V_2/M_1 = 0.292 M times 100 mL/1.0 M V_1 = 29.2 mL of 1.0 MH_3 PO_4 Molecular mass of KH_2 PO_4 = 136.086 gmol^-1 Molarity = mass of KH_2 PO_4/Moleular mass of KH_2 PO_4 times volume of buffer mass of KH_2 PO_4 = Molarity times Moleular mass of KH_2 PO_4 times volme of buffer mass of KH_2 PO_4 = 0.208 M times 136.086 gmol^-1 times 0.10L mass of KH_2 PO_4 = 2.830 g