Post-lab 7 Weak acid buffer-solutions

Post-lab 7 Weak acid buffer-solutions - choudhury (fac489)...

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Unformatted text preview: choudhury (fac489) – Post-lab 7 Weak acid buffer – lyon – (51155) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points All components are present in 0.10 M concen- trations. I) HCN and NaCN II) NH 3 and NH 4 Cl III) HNO 3 and NH 4 NO 3 IV) HClO 3 and NaClO 3 Which will give buffer solutions? 1. III and IV only 2. I and III only 3. II, III and IV only 4. I, III and IV only 5. I and II only correct Explanation: Buffers are formed in one of two ways, by combining a weak acid and its conjugate base or by combining a weak base and its conjugate acid. HNO 3 and HClO 3 are both strong acids and cannot be used to make effective buffer solutions. HCN is a weak acid and NaCN is the salt of its conjugate base, CN − . NH 3 is a weak base and NH 4 Cl is the salt of its conjugate acid, NH + 4 . Therefore 1 and 2 can be used to make effective buffer solutions. 002 10.0 points Which response identifies the statements true of buffer solutions? l) A buffer solution could consist of equal concentrations of ammonia and ammo- nium bromide. 2) A buffer solution could consist of equal concentrations of perchloric acid (HClO 4 ) and sodium perchlorate. 3) A buffer solution will change only slightly in pH upon addition of small amounts of acid or base. 4) In a buffer solution containing benzoic acid (C 6 H 5 COOH) and sodium benzoate (NaC 6 H 5 COO) the species that reacts with added hydroxide ion is the benzoate ion. 1. 1, 4 2. Another combination 3. 2, 3, 4 4. 2, 3 5. 1, 3 correct Explanation: HA + H 2 O ⇀ ↽ A − + H 3 O + B + H 2 O ⇀ ↽ BH + + OH − NH + 4 is the conjugate acid of the weak base NH 3 ; C 6 H 5 COO − is the conjugate base of the weak acid C 6 H 5 COOH. HClO 4 is a strong acid. OH − ions introduced into an acid buffer system removes the H 3 O + . Buffers are composed of comparative amounts of a weak acid/base and its conjugate and allows only a small change in pH when an acid or base is added. 003 10.0 points A solution of 50 mL of 0.3 M acetic acid is titrated with 75 mL of 0.2 M NaOH. What is the pH of the resulting solution? K a for acetic acid is 1 . 8 × 10 − 5 ? 1. 8.91 correct 2. 5.1 3. 12.1 4. 7.00 Explanation: Acetic acid reacts with sodium hydroxide: HAc + NaOH → NaAc + H 2 O (0 . 05 L HAc) × . 3 mol HA 1 L HA = 0 . 015 mol HA choudhury (fac489) – Post-lab 7 Weak acid buffer – lyon – (51155) 2 (0 . 075 L NaOH) × . 2 mol NaOH 1 L NaOH = 0 . 015 mol NaOH All HAc and NaOH reacts to give 0.015 mol Na + and 0.015 mol Ac − . The Na + has no effect on pH. Ac − is a weak base; the initial concentra- tion of Ac − is [Ac − ] = . 015 mol . 125 L = 0 . 12 M We set up an ICE table with molarities: Ac − + H 2 O ⇀ ↽ HAc + OH − I . 12 C- x + x + x E 0 . 12- x x x K b = [HAc] [OH − ] [Ac − ] = K w K a = 1 × 10 − 14 1 . 8 × 10 − 5 = 5 . 55556 × 10 − 10 Assuming x is small, K b = x · x . 12- x...
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This note was uploaded on 12/13/2011 for the course CH 204 taught by Professor Leytner during the Fall '08 term at University of Texas at Austin.

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Post-lab 7 Weak acid buffer-solutions - choudhury (fac489)...

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