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

Question

When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a butter 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^+, 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 = pK_a + log [A^]/[HA] What is the pH of a buffer prepared by adding 0.506 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 Part A? 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 Part A? Assume no volume change on the addition of the base. Express the pH numerically to three decimal places.

Explanation / Answer

Part-A:

Concentration of weak acid, HA, [HA] = moles of HA / Vt = 0.506 mol / 2.00L = 0.253 M

Concentration of conjugate base, A-, [A-] = moles of NaA / Vt = 0.305 mol / 2.00L = 0.1525 M

Now the pH of the buffer solution can be calculated for Hendersen equation

pH = pKa + log[A-] / [HA]

=> pH = - log(5.66x10-7) + log(0.1525 / 0.253) = 6.03 (answer)

Part-B:

When 0.150 mol of HCl is added it will react with 0.150 mol of A- to form 0.150 mol HA.

------------ HCl + A-  ------ > HA + Cl-(aq)

init mol: 0.150, 0.305 ------- 0.506

change: - 0.150, - 0.150, --- + 0.150

eqm.mol: 0 mol, 0155 mol, 0.656 mol

Since volume doesn't change,

[A-] / [HA] = moles of A- / moles of HA = 0.155 / 0.656

Now applying Hendersen equation

pH = pKa + log[A-] / [HA]

=> pH = - log(5.66x10-7) + log(0.155 / 0.656) = 5.62 (answer)

Part-C:

When 0.195 mol of NaOH is added it will react with 0.195 mol of HA to form 0.195 mol A-.

------------ NaOH + HA ------ > A- + Na+(aq)

init mol: 0.195 , 0.506 ----- 0.305

change: - 0.195, - 0.195, --- + 0.195

eqm.mol: 0 mol, 0.311 mol, 0.500 mol

Since volume doesn't change,

[A-] / [HA] = moles of A- / moles of HA = 0.500 / 0.311

Now applying Hendersen equation

pH = pKa + log[A-] / [HA]

=> pH = - log(5.66x10-7) + log(0.500 /0.311) = 6.45 (answer)

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