Question Which one of the following procedures would result in a buffer with pH close to 5.0? Ka(CH3COOH) = 1.8 × 10-5 Kb(NH3) = 1.8 × 10-5 Mixing 1 mol NH3 and 0.5 mol HCl in a 1.0 L aqueous solution. (Your answer) Mixing 1 mol CH3COOH and 1 mol NaOH in a 1.0 L aqueous solution. Mixing 1 mol CH3COOH and 0.5 mol NaOH in a 1.0 L aqueous solution. (Correct answer)

ND5HNM The Asker · Chemistry

Which one of the following procedures would result in a buffer with pH close to 5.0? Ka(CH3COOH) = 1.8 × 10-5 Kb(NH3) = 1.8 × 10-5 Mixing 1 mol NH3 and 0.5 mol HCl in a 1.0 L aqueous solution. (Your answer) Mixing 1 mol CH3COOH and 1 mol NaOH in a 1.0 L aqueous solution. Mixing 1 mol CH3COOH and 0.5 mol NaOH in a 1.0 L aqueous solution. (Correct answer) Mixing 1 mol CH3COOH and 1 mol NH3 in a 1.0 L aqueous solution. Mixing 1 mol NH3 and 1 mol HCl in a 1.0 L aqueous solution.

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Soln:- Given data are as follows:Ka of CH_(3)COOH=1.8 xx10^(-5),K_(b) af NH_(3)=1.8 xx10^(-5)80pKaggCH_(3)COOH=4.745,quadpk_(b) of NH_(3)=4.745then as we know{:[P_(Ka)+P_(K_(b))=14],[" then, "P_(K_(a))=14-P_(K_(b))=9.525]:}Here we have to pre pare of buffer of rho H=5.0 PH of Buffer soln is given by.pH=pka+log(([" Basis "])/([" Acid "])),quadpoH=pk_(b)+log(([" consugahe acid "]) ... See the full answer