A solution of a weak acid (HA) and its conjugate base (A-), HA(aq)-->H+(aq)+A-(a

Question

A solution of a weak acid (HA) and its conjugate base (A-), HA(aq)-->H+(aq)+A-(aq) is called a buffer and will resist a change in pH. If acid is added, the reaction shifts to consume the added H+, forming more HA. When base is added, the base will react with H+, reducing its concentration. The reaction then shifts to replace H+ through the dissociation of HA into H+ and A-. In both instances, [H+] tends to remain constant. The pH of a buffer is calculated by using the Henderson-Hasselbalch equation: pH=pKa+log[A-][HA]

Explanation / Answer

(A) pH = pKa + log([NaA]/[HA])

= -log Ka + log(moles of NaA/moles of HA)

= -log(5.66 x 10^(-7)) + log(0.609/0.405)

= 6.42


(B) NaA + HCl => NaCl + HA

Moles of HA = 0.405 + 0.150 = 0.555 mol

Moles of NaA = 0.609 - 0.150 = 0.459 mol


pH = pKa + log([NaA]/[HA])

= -log Ka + log(moles of NaA/moles of HA)

= -log(5.66 x 10^(-7)) + log(0.459/0.555)

= 6.16


(C) HA + NaOH => NaA + H2O

Moles of HA = 0.405 - 0.195 = 0.210 mol

Moles of NaA = 0.609 + 0.195 = 0.804 mol


pH = pKa + log([NaA]/[HA])

= -log Ka + log(moles of NaA/moles of HA)

= -log(5.66 x 10^(-7)) + log(0.804/0.210)

= 6.83

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