T2%20Solutions - You are writing VERSION 1 of this test....

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1 Continued on next page… You are writing VERSION 1 of this test. Make sure you have correctly entered your version number (“1”) in the correct column on your scan sheet (see p. 2 for details). ____ 1. What is the pH of a 0.0242 M solution of HNO 3 ? a. 4.214 b. 1.62 Part Mark c. 2.11 d. 3.18 e. 1.616 ____ 2. Which of the following will create a basic solution in water? i. MgF 2 ii. NH 4 Cl iii. LiBr iv. NaCHOO v. K 3 PO 4 a. i, iv, v b. ii, iii, iv c. i, iv – Part Mark d. ii, iv e. iii Section #1 – These questions are worth two marks each. Strong acid therefore it fully dissociates. [H 3 O + ] = 0.0242 pH = -log(0.0242) pH = 1.616 Note the sig. figs. with logs. i. Yes! F - is conjugate of weak acid (HF) ii. No! NH 4 + is conjugate of a weak base (NH 3 ) iii. No! Both are conjugates of a strong acid & base iv. Yes! CHOO - is the conjugate of weak acid (CHOOH) v. Yes! PO 4 3- is the conjugate of a weak acid (HPO 4 2- )
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2 Continued on next page… ____ 3. Which of the following statements about the accompanying reaction is correct ? a. H 2 O + SO 3 H 2 SO 4 H 2 O is a Lewis acid b. NH 4 + + H 2 O NH 3 + H 3 O + The solution will be alkaline c. SnCl 4 + 2Cl - [SnCl 6 ] 2- Cl - is a Lewis base d. OH - + CO 2 HCO 3 - CO 2 is a Lewis base e. CH 3 NH 2 + H 2 O CH 3 NH 3 + + OH - H 2 O is a Bronsted-Lowry base ____ 4. What is the pH of a 0.140M NaOCN solution? ( K a HOCN = 3.5 x 10 -4 ) a. 7.83 b. 8.30 c. 9.53 d. 12.70 e. 6.21 ____ 5. Which of the following acids has the strongest conjugate base ? a. H 2 C 6 H 5 O 7 - K a = 4.0 x 10 -7 b. HOC 6 H 5 K a = 1.0 x 10 -10 c. CH 3 COOH K a = 1.8 x 10 -5 d. HSO 4 - K a = 1.1 x 10 -2 e. (CH 3 ) 3 NH + K a = 1.6 x 10 -10 a. False. H 2 O is a Lewis base b. False. H 3 O + is produced. The solution is acidic c. True d. False. CO 2 accepts electrons from OH - (Lewis acid) e. False. Water donates a proton (B.L. acid) OCN - + H 2 O HOCN + OH - I 0.140 ----- 0 0 C -x ----- +x +x E 0.140 – x ----- x x Rule of 100 can simplify 0.140 – x to 0.140 2.9 × 10 11 = x 2 0.140 x = 2.0 × 10 6 pOH = log(2.0 × 10 6 ) pOH = 5.69897 pH = 14 5.69897 pH = 8.30 The weaker the acid, the stronger the conjugate base, therefore, look for the acid with the lowest K a
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3 Continued on next page… ____ 6. What is the percent ionization of 0.250M ascorbic acid? (p K a = 4.17) a. 9.1% b. 2.1% c. 1.6% d. 6.7% e. 7.8% ____ 7. Copper metal is currently $3.00 / lb, while zinc metal is $1.00 / lb. If you were given 1000 lb of CuO and unlimited, free sulfuric acid, how much money could you make by producing copper metal with zinc? a. $ 1570 b. $ 3880 c. $ 3010 d. $ 1150 e. $ 789 HAbs + H 2 O Abs - + H 3 O + I 0.250 ----- 0 0 C -x ----- +x +x E 0.250 – x x x 6.761 × 10 5 = x 2 0.250 x = 4.111 × 10 3 4.111 × 10 3 0.250 × 100 = 1.6% CuO(s) + H 2 SO 4 (aq) CuSO 4 (aq) + H 2 O(l) CuSO 4 (aq) + Zn(s) ZnSO 4 (aq) + Cu(s) Note all 1:1 ratios Molar mass of CuO = 79.545 g/mol % comp of copper in CuO = 63.546 / 79.545 x 100 = 79.89%
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T2%20Solutions - You are writing VERSION 1 of this test....

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