A) Find the pH of a 1.00-L aqueous solution prepared with 10.00 g of imidazole (

Question

A) Find the pH of a 1.00-L aqueous solution prepared with 10.00 g of imidazole (FM 68.08 g/mol) and 17.50 g of imidazole chloride ( FM 104.54 g/mol). (use table provided)

B) If 25.00 mL of 0.500 M NaOH solution are added to the buffer soltion in part A, what is the pH now?

TABLE 8-2 Structures and pKa values for common buffers b,cd Structure CH,CO,H N-2-Acetamidoiminodliacetic acid. H2NCCH2NH CH,CO, H (ADA) N-Tris(hydroxymethyl)methylglycine CHoCH acNH2CH2CO2H Phosphoric acid H3PO4 NN-Bis(2-hydroxyethyl glycine (HOCH,CH NHCH CO2H (BICINE) (see above) Piperazine-N,N -bis(2-ethanesulfonic NH CH, CH,SO acid (PIPES) OH Citric acid CO Glycylglycine HNCH, CH CO2H Piperazine-N, N'-bis(3-propanesulfonic osoCHON (CH2 So acid (PIPPS) Piperazine-NN -bis(4-butanesulfonic -o,s CH AN acid) (IPBS) NIN -Diethylpiperazine CH2CH3-2C1 dihydrochloride (DEPP-2HCl) Citric acid (see above) Acetic acid CH3CO2 NIN-Diethylethylenediamine-N,N'- CH3CH2 CH2CH3 DESPEN) 2-(N-Morpholino ethanesulfonic acid o So HCH,CH (MES) Citric acid (see above) N,N,N,N -Tetraethylethylenediamine EUNHCH,CH,HNEtn 201 dihydrochloride TEEN 2HC1) 1,3-Bis[tris(hydroxymethyl)methyl ,CNH (CH2,NH2, 2C1 amino] propane hydrochloride (HOCH (BIS-TRIS propane-2Hci) (see above) 0.1 M (CO2H) 1.59 2.02 (CO2H) 2.15 (pKi) 1.92 2.23 (CO2H) 2.48 (CO2H) 2.31 (pKI) 2.67 3.13 (pK1) 2.90 3.14 (CO2H) 3.11 (pKI) 3.79 (pKI) 4.29 (pKI) 4.48 4.76 (pK2) 4.35 4.76 4.62. (pKI) 5.62. 6.27 6.06 6.40 (pK3) 5.70 (pKI) 6.58 6.65 (pKi) 6.84 (NH) 6.67 Formula mass 90.15 179.17 98.00 163.17 90.15 302.37 92.12 132.12 330.42 358.47 215.16 92.12 60.05 360.49 195.24 92.12 245.23 355.26 90.15 A(pKa)/AT 0.003 0.005 0.002 0.000 0.001 0.000 0.009 0.002 0.007

Explanation / Answer

Moles of imidazole = 10gm/68.08g/mol = 0.147 moles

moles of imidazolinium chloride = 17.5fm/104.54 = 0.167 moles

pkb of imidazole =7.01

pOH = pKb + log [salt/base]

        = 7.01 + log [0.167/0.147] = 7.065

pH = 14- 7.065 = 6.93

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moles of NaOH added =0.5 M * 0.025 L = 0.0125 moles

moles of inidazole present after adding NaOH = 0.147 + 0.0125 =0.1595

moles of inidazolium salt present = 0.167-0.0125 = 0.1545

pOH = 7.01 + log [0.1595/0.1545] = 7.01

pH = 14- 7.01 = 6.99

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