Table 10.2 continued pK Cougate base 4.75 CH3COO Acid name Acid formula Conjugat
ID: 925586 • Letter: T
Question
Table 10.2 continued pK Cougate base 4.75 CH3COO Acid name Acid formula Conjugate base name acetic acid CH3COOH acetate ion 9.25 dihydrogen citrate ion 3H,o(COOH)2COO 4.77 C3Hso(CoOH)(CO hydrogen citrate ion 9.23 butanoic aicd propanoic acid pyndinium ion carbonic acid C3H,COOH C2Hs COOH CSH,NH HOCOOH 4.82 C3H,COO 4.87 C2H,COO 5.25 CSH,N 6.37 HOcoo butanoate ion propanoate ion pyridine 9.13 hydrogen carbonate ion 7.63 hydrogen citrate ionC3H,o(COOH)(COO2 6.40 C3Ho(CO0) 7.60 6.81 7.19 SO 7.20 HPO 7.53 CIO 9.25 NH3 9.31CIN 10.00 CHsO 1025 CO3 10.49 (C2Hs)2NH 10.64 CH,NH 10.71 C3H,NH2 10.77 | (CH),NH 10.87 C2HsNH2 11.01 (C2H5)3N 12.38 Po citrate ion sulfite ion hydrogen phosphate on .80 hypochlorite ion ammonia cyanide ion phenolate ion carbonate ion diethylamine methylamine propylamine dimethylamine ethylamine triethylamine phosphate ion hydrogen sulfite ion dihydrogen phosphate ion H2PO HCIO hypochlorous acid ammonium ion hydrocyanic acid phenol hydrogen carbonate ion HOCoo HCN 4.69 4.00 3.75 C&H;,OH diethylammonium ionC2Hs)2NH2 CH,NH C3HNH methylammoniu m ion propylammonium ion 3.29 dimethylammonium ion (CH3)2NH 3.23 ethylammonium ion C2H,NH triethylammonium on C2H5)NH hydrogen phosphate ion HPO 2.99 1.62Explanation / Answer
moles of butanoic acid = moles of butanoate + moles of butanoic acid in solution
let x amount of base is added then,
moles of butanoic acid = 0.072 x 0.78 = 0.05616 mols
pH = pKa + log([base]/[acid])
4.76 = 4.82 + log(x/0.05616-x)
0.050 - 0.871x = x
x = 0.027 mols
So the moles of butanoate present would be = 0.027 mols
1 mole of Sr(OH)2 gives 2 moles of butanoate
thus, moles of Sr(OH)2 required = 0.027/2 = 0.0135 mols
Volume of Sr(OH)2 to be added = 0.0135/0.360 = 0.0375 L
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