Question

How many grams of sodium acetate, NaC₂H₃O₂, would have to be added to 1.0 L of 0.90 M acetic acid (pK_a = 4.74) to make the solution a buffer for pH 4.00?

Answer 1

Na GHO =CH₂COona CH₃COona - CH₃COO- that when a salt dissociates, negative ion is called conjugate base of acid Here conjugate base - [CHgcoo]. [bass [CH₂ coo- ] = { base] Now, acetic acid & CH₃COOH =ęHUO2 [acid} = [CH₃COOH] = 0.90M pka - 4.74 pH=4 pH = pka + 2log (base) To lacid] 4 = 4.74 + 6g (base] ] log (CH₂ coon] Jio facid] 010 0.90 ) = -0.74 Ссна соо) - о-о-ч [CH₂ 600-] = (0.9) (0.18197). [CH3C00] = 0.1637M CH₃COONa imol Volume =L ; CH3000 that mol Mohnity = molle in IM IM

[CH₃COONa] = [CH₂ coo- ] [CH₂ coona] = 0.1637 Given volume = IL Molasity a moles volume in i 0.1637 - moles of CHCOona moles of CH₃COONa = 0.1637 mol Molar mass of CH₂ coona = 82.03 glool Moles = mass Molar mass Mass of CH₃COONa = (0.16 37 ml) (82.03 9 ) = 13.428 g mal Mass of sodium acetate = 13.43 g