Question

Part A: What is the pH of a buffer prepared by adding 0.708 mol of the weak acid HA to 0.406 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7.

Part B: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.

Part C: 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.

Answer 1

- [H] CAT Parta: Acid buffer of weak acid HA and Naa Weak acid dissociated and attained as- HA => H+ + A, Ka=b! - [HA PH of the buffer solution is given by → pH = pka + long та [HA] Here, given [HA] = 0-708 mol Total volume = 2L Na Al = 0.406 mol ka = 5066 X10-7 Þka = -log ka = -109 (5-66X107 .: BBQ = 6.25 Now, pH = pka + log wat = 6-25 + log 9: 406 [PH z 6.00] is pH of the buffer solution = 6.00 Part B : after addition of 0.150 mol of Hel to the buffer, the following changes takes Place → Hei fadded] + Nad butter) -> HA + NaCl Now, [HA] and [NaA] Changes, as HA is formed and NaA is reacted: So, [14] = 10.708 +0.150) mol = 0.858 mol [NaA] = 0.106 – 0.150) mol = 0.256 mol

NOW DH = bko +100 [naA] = 6.25 + log 0-256 [PH = 5.73 = 5.73 After aditition at 0.150 mol of Hal to the butter pH becomes 5.73 : Part . After addition of 0.195 mol NaOH to the bufter of ÞH = 6.00, following ehange oecurso Nach ladded) + HA (butter → NG A + H20 Here, Na A Es formed and HA is reacted, so [141] and [NaA] Changes and it becomes → [HA] = (0.708 - 0.195) mol = 0.573 mol [NaA] = 10.406 + 0.195 mol = 0.601 mo! Now, pH = pka + ing NGAT [HA] = 6.25 + log o bobl = 6-31 [ÞH = 6:31 1 After addition of 8.195 mol of Naot to the butter the pH becomes 6-31.
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