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When a solution contains a weak acid and its conjugate base or a weak base and i

ID: 997470 • Letter: W

Question

When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. Buffers resist change in pH following the addition of acid or base. A buffer solution prepared from a weak acid (HA) and its conjugate base (A) is represented as HA(aq) H^+ (aq) + A (aq) The buffer will follow Le Chatelier's principle. If acid is added, the reaction shifts to consume the added H^1. forming more HA. When base is added, the base will react with H^1. reducing its concentration. The reaction then shifts to replace H^1 through the dissociation of HA into H^1 and A. In both instances. [H^1] tends to remain constant. The pH of a buffer is calculated by using the Henderson-Hasselbalch equation: pH = pK_a + log [A^-]/[HA] What is the pH of a buffer prepared by adding 0.708 mol of the weak acid HA to 0.305 mol of NaA in 2.00 L of solution? The dissociation constant K_a of HA is 5.66 times 10^-7 Express the pH numerically to three decimal places. What is the pH after 0.150 mol of HCl is added to the buffer from ? Assume no volume change on the addition of the acid. Express the pH numerically to three decimal places. What is the pH after 0.195 mol of NaOH is added to the buffer from ? Assume no volume change on the addition of the base. Express the pH numerically to three decimal places.

Explanation / Answer

PH = Pka + log[NaA]/[HA]

Ka = 5.66*10-7

Pka = -logka

       = -log5.66*10-7

     = 6.247

PH   = 6.247 + log0.305/0.708

     = 6.247-0.3657   = 5.8813 >>>> answer

part B

By addition of 0.15 moles of HCl

no of moles of HA = 0.708+ 0.15 = 0.858 moles

no of moles of NaA   = 0.305-0.15 = 0.155 moles

PH = Pka + log[NaA]/[HA]

      = 6.247 + log0.155/0.858

    = 6.247-0.7431 = 5.5039

part c

By addition of 0.195 moles of NaOH

no of moles of HA = 0.708-0.195 =0.513 moles

no of moles of NaA   = 0.305 + 0.195 = 0.5 moles

PH = Pka + log[NaA]/[HA]

      = 6.247 + log0.5/0.513

     = 6.247 -0.01114 = 6.2358 >>>> answer

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