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2 hem 16 Exam arch 6, 010 Ellis NONE OF THE UTPE HOE PROBES REQURES EXTENSVE OR TE OSUNG AUATONS. F YOUR ETHOD REQURES EXTENSTVE AUATONS T S ETHER WRONG OR HARDER THAN WHAT S REQURED. 1. A 0.1 solution of each acid in the table is prepared. Which has the HGHEST ph. areful. a. Acetic b. Fluoroacetic c. Formic d. Hydrocyanic e. Hydrofluoric The highest ph means the least acidic or the one with the smallest K. (ost acidic or lowest ph has largest K.) a a..5 g NaOH is diluted to 5.0. What is the ph? a b. 1.1 W.. NaOH= = 40 g/ mol c mol.5g d. 1.7 mol 40g = = = poh = log OH log ( 0.015) 1.90 e = = ph = 14 poh = m of 0.45 HNO is mixed with 00. m of 0.55 HN? What is the ph? a b. 0.6 c. 0.5 d e. 0.8 ( ) ( ) Strong acid Hl + weak acid HN ; strong acid determines the ph + 100m + H = 0.45 = 0.15 = ph = log H = log ( 0.15) = m m of 0.0 HNO is mixed with 40. m of 0.0 Hl? What is the ph? a Strong acid + strong acid b mol 60m 0.0mol 40m c. 0.9 mmol HNO = x = 18 mmol; mmol Hl = x = 8mmol 1 1 d mmol mol + H = = 0.6 = 0.6 ; ph = log H = log( 0.6) = 0.59 e m

3 hem 16 Exam arch 6, 010 Ellis m of 0.6 Hl is mixed with 80. m of 0.10 KOH. What is the ph? a b c d e. >10 Acid-base neutralization: + 0.6mol 15m 0.1mol 80m mmol H = x = 9. mmol; mmol OH = x = 8mmol H aq OH aq H O aq ( ) + ( ) ( ) E mmol H = = ; 95m ph = log ( 0.017) = What is the hydrogen ion concentration of a 0.77 HF? K a = 6.76 x a. < 0.01 b c. 0.0 d e ( ) + ( ) ( ) + ( ) HF aq H O aq HO aq N ag ~0 0 -x +x +x E 0.77-x x x + HO F x = 10 4 Ka; = 4.79x10 ; x = 6.76x10 ( 0.77 x) 0.0 HF 0.77 x [ ] (ethod of successive approximations used; first iteration: 0.0; converged 0.0.) 7. A solution of a weak acid, HB, has a ph of.0. What is K a? a. 1.8x10-4 b. 1.x10-4 c..5x10-7 d. >10 - e. <10-8 = = = + ph.0 4 H x10 [ HB] + ( ) + ( ) ( ) + ( ) HB aq H O aq H O aq B ag ~ x x x E x x1 4 ( ) + 4 HO B x K a = ; = 1.8x Which of the following ions is neutral? a. Al + + d. NH 4 b. Br - e. Br OO - c. F - - Br is a neutral ion conjugate base to a strong acid - F and BrOO Al + NH + 4 ( -- HBr). are conjugate bases to weak acids andarebasic., like most transition metal ions, is acidic. is the conjugate acid to a weak base and is acidic.

4 hem 16 Exam arch 6, 010 Ellis 9. H B (B = conjugate base) is a diprotic acid. A solution of 0. KHB is prepared. s it acidic, basic, or neutral? + 6 HBaq + HOaq HO aq + HB aq K = 510 x ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) a1 HB aq + H O aq H O aq + B aq K = x a a. Neutral because K a and K b for HB - = x10-9. b. Acidic because K + is an acidic ion. c. Acidic because K a for HB - = 5x10-6 > K b for HB - = x10-9 d. Basic because K a for HB - = x10-9 < K b for HB - = 5x Ka for HB as acid is Ka = x 10 and 14 Kw 10 Kb as base is = = x 6 K 510 x a m of 0.0 acetic acid is added to 00 m of 0.40 sodium acetate. What is the ph? K a = 1.74 x a. 4. The solution consists of a weak acid and its conjugate base: use buffer equation. b c d e mol pka = log ( 1.74x10 ) = 4.76; mmol acid = 100m = 0mmol 0.4mol B 80 mmol base 00m 80 mmol; ph pka log = = = + = log [ HB] = Hydoxylanime (NH OH) is a weak base like ammonia. K b = 1.07x10-8 and pk b = t can be used to prepare a buffer solution whose ph is 6.0. f the NH OH concentration is 0.400, what is the required NH OH + concentration? areful! a. < 0.1 b c d e. > 5 B pka = Kw pkb = = 6.0; ph = pka + log [ HB] x 6.0 = log ; 0.7 = log ; log =0.7 x x NH OH = x = = Buffers neutralize both acids and bases. When HBr is added to a buffer solution consisting of Sodium cyanide (NaN) and hydrocyanic acid (HN), which reaction (when completed) shows the neutralization of the acid? a. H 0 + (aq) +N - (aq) b. H 0 + (aq) + Na + (aq) c. H 0 + (aq) + Br - (aq) d. H 0 + (aq) + HN(aq) e. None of the above

5 hem 16 Exam arch 6, 010 Ellis m of 0.0 Hl is added to 60.0 m of 0.75 NaH OO. What is the ph of the resulting solution? (K a = 1.74 x 10-5 for acetic acid.) a b. 5. c d e E pk ( ) + ( ) HO( l) + HOOH( aq) + 0.mol 155m = 1mmol 45mmol 0 1 mmol 1 mmol + 1mmol a HO aq HOO aq 0 mmol 14 mmol 1mmol 5 H OO 14 = log ( x ) = 4.76; ph = pka + log = log = 4.41 HOOH What is the ph of 0.55 NaF. K a for HF is 6.76 x K 10 HO l F aq OH aq FH aq K x a. 0.6 b c. 1.7 x + x + x E 0.55 x x x d x e = Kb; x = 1.48x10 ( 0.55x) 1.48x10 ( ) =.85x10 55 x 14 w 11 () + ( ) ( ) + ( ) b = = = Ka 6.76x10 6 ( ) (.85x10 ) poh =log OH = log = 5.54; ph = = n which of the following reactions does the underlined compound act as a either a BrØnsted- owry or ewis base? (Note, and are different reactions.). HO + H O H O + OH. HO + H O O + H O + ( ) ( ) Base because it accepts proton from water. Base because it accepts proton from HO.. PH+ B OH PH B OH where the P is bonded to the B ewis acid because B accepts lone pair of electrons on P. (a) only (b) only (c) only (d) and (e),, and 16. Which of the following solutions is not a buffer solution? a. 0.1 acetic acid and 0.5 conjugate base. b. 0. Fe(H O) + 6 and 0. Fe(H O) 5 OH + d. 0. N(H ) and 1.4 NH(H ) + e H + - and 0.60 NO

6 hem 16 Exam arch 6, 010 Ellis PROBES (1 PONTS EAH). (a) The magnitude of the K sp of Fe(OH) is 1.0 x at room temperature. ( PTS) Write the solubility reaction or the solubility product expression (K sp =?): + ( ) ( ) " ( ) + ( ) Fe OH s Fe aq OH aq or + Fe OH K sp = (b) (5 PTS) What is the solubility of Fe(OH) in pure water at room temperature in mol/? USE ORRET NUBER OF SGNFANT FGURES. E + ( ) ( ) " ( ) + ( ) Fe OH s Fe aq OH aq excess 0 0 -s +s +s excess s s ()( s) 15 1/ x10 6 s = 1x10 ; s = = 6.x10 mol/ 4 (c) (5 PTS) What is the solubility of Fe(OH) in a solution buffered to a ph of 11 at 5º? poh = 14 11= ; OH = 10 = 1.00x10 poh E + ( ) ( )" ( ) + ( ) Fe OH s Fe aq OH aq excess s +s +0 (buffered solution!) excess s x10 9 s10 ( ) = x ; s = = x mol/ 6 10

7 hem 16 Exam arch 6, 010 Ellis. 5 ml of a KN are titrated with 0.50 Hl. (a) ( PTS) What is the neutralization reaction? ( ) + N ( aq) H O ( aq) + HN ( aq) ( ) or Hl ( aq) + N ( aq) HN ( aq) HO aq better + (b) ( PTS) t takes.0 m of the acid to reach the equivalence point (to neutralize all of the base). What was the initial concentration of KN? 0.5molHl m 1mol KN 5.5mmol KN x x = 5.5 mmol KN; = KN 1 1mol Hl 5m (c) ( PTS) After 11.0 m of acid are added to the original solution the ph is 9.. What is K b for N -? Hint, what is N - converted to? After 11.0 m of acid are added, we are at the midpoint (half the N is converted to HN). So, N ph = pka + log = pka = 9.; pkb = = 4.68; Kb = 10 =.09x10 [ HN ] (d) ( PTS) What is the ph after 0 m of acid are added to the original 5 m of KN? 0.5mol 0m 0.157mol 5m mmol acid = x = 7.5 mmol; mmol base = x = 5.5mmol 1 1 E ( ) + ( ) ( ) + ( ) + HO aq N aq HO aq HN aq Some strong acid and some weak acid. Use strong acid approximation. ( ) m ph = log H + =log mol = m

8 hem 16 Exam arch 6, 010 Ellis. Pot-pourri (a) ( PTS) A solution of Pb(H OO) is added to a solution of Na. A yellow precipitate forms. What is it? Pb + (b) ( PTS) ( ) ( ) ( ) HO l H aq OH aq H + > 0 7. What is the ph of pure water just above freezing? Explain why.. At room temperature the ph of pure water is For an endothermic process, decreasing the temperature means fewer water molecules have sufficient energy to dissociate (fewer beads have sufficient energy to jump to the higher level). That means fewer hydrogen ions. So, the ph increases. (c) (4 PTS) The solubility product constant, K sp, of a(oh) is 5.0 x m of a(no ) is added to 40. m of a buffer solution having a ph of What is Q? areful. 15 ph = ( ) = 0.07 = ; = 14 1 = ; = 10 = a a NO x poh OH + ( ) ( ) ( ) + ( ) a OH s a aq OH aq Q ( 1 ) + = a OH = x = x10 6 (d) ( PTS) n the last problem, does a precipitate form? Why? (No credit without proper justification--just a few words.) No, because Q< K sp.