Unformatted text preview: 4.5 = 4.745  log10(0.10/[NaC2H3O2]) 0.245 = log10(0.10/[NaC2H3O2]) 0.245 = log10(0.10) – log10([NaC2H3O2]) log10([NaC2H3O2]) = 1 – 0.245 =  1.245 [NaC2H3O2] = 10^(1.245) = 0.056885 ______________________________________. .. n = C*V à moles = concentration (mol/L) * Volume (L) n = = 0.056885 mol/L * 0.1L = 5.6885 x 103 ______________________________________. .. m = n*M à mass = mole * molecular weight molecular weight of NaC2H3O2 = 82.024 …. . (from your periodic table) therefore m = 82.024 x 5.6885 x 103 = 0.4666 g ANSWER = 0.4666 g...
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 Spring '07
 PottsSantone
 conjugate base, 10 m, 0.1L, Henderson Hesselbach equation, 0.4666 g

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