Chapter 04 Molarity-Dilution-Stoichiometry with Solutions Worksheet (KEY)

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1 Chapter 04 Molarity-Dilution-Stoichiometry with Solutions Worksheet (KEY) 1. A person sufferin from hyponatremia has a sodium ion concentration in the blood of M and a total blood volume of 4.6. What mass of sodium chloride would need to be added to the blood to brin the sodium ion concentration up to M, assumin no chane in blood volume? M = n V ; n = M V m NaCl () = Na 4.6 ( ) = Na Initially Na 4.6 ( ) = Na Required Na Na = Na to add Na 1 NaCl NaCl ( ) ( ) = NaCl 1 Na 1 NaCl 2. Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinklin sodium bicarbonate on it and then moppin up the resultant solution. The sodium bicarbonate reacts with sulfuric acid as follows: 2 NaHCO 3 (s) H 2 SO 4 (aq) Na 2 SO 4 (aq) 2 H 2 O (l) 2 CO 2 () Sodium bicarbonate is added until the fizzin due to the formation of CO 2 () stops. If 27 m of 6.0 M H 2SO 4 was spilled, what is the minimum mass of NaHCO 3 that must be added to the spill to neutralize the acid? M = n V ; n = M V m NaHCO3 () = m ( 6.0 H 2SO m H 2 SO 4 ) = H 2 SO 4 MM NaHCO3 = ( ) = H 2 SO 4 ( 2 NaHCO 3 1 H 2 SO ) ( NaHCO 3 ) = NaHCO 4 1 NaHCO 3 3

2 3. Two solutions of ammonium phosphate are mixed; for the first V = 35.0 m and M = 1.00; for the second V = 60.0 m and M = The solution obtained by this mixin is then heated to evaporate the water until the total volume is 50.0 m. What is the arity of the ions present in the final solution? M = n V ; n = M V 1.00 n 1 = 35.0 m ( 1000 m ) = ; n = 60.0 m ( ) = m M Cation = 4.26 M Anion = 1.42 n T = n 1 n 2 = = (NH 4 ) 3 PO 4 M = (NH 4) 3 PO m ( ) = 1.42 M (NH 50.0 m 1 4 ) 3 PO 4 M NH 4 = 1.42 (NH 4) 3 PO 4 M PO 4 3 = 1.42 (NH 4) 3 PO 4 3 NH 4 ( ) = 4.26 NH 4 1 (NH 4 ) 3 PO PO 4 ( ) = 1.42 PO 4 1 (NH 4 ) 3 PO By titration, 15.0 m of M sodium hydroxide is needed to neutralize a sample of a monoprotic oranic acid. a. What is the ar mass of the acid if it is monoprotic? b. An elemental analysis of the acid indicates that it is composed of 5.89% H, 70.6% C and 23.5% O by mass. What is its ecular formula? Since the acid is monoprotic, the es of NaOH are the same as the es of acid NaOH 15.0 m ( ) = NaOH = Oranic Acid 1000 m MM Oranic Acid = = H 5.89 H ( ) = H 1 C 70.6 C ( ) = C ; 23.5 O ( 1 ) ) = O H O H = 3.98 O ; C C = O O EF: C 4 H 4 O ; MMEF = ( ) ( ) = n = = ; MF = (C 4 H 4 O) 2 = C 8 H 8 O 2 MM of Oranic Acid = 136 / Molecular Formula: C 8 H 8 O 2

3 m of a M old (III) nitrate solution are mixed with 47.2 m of a M iron (III) nitrate solution. Once the mixture is obtained, heat is applied until half of the water boils away. Give the final concentration of the nitrate ion. n NO3 from Au(NO 3 ) 3 = 80.5 m ( Au(NO 3) m n NO3 from Fe(NO 3 ) 3 = 47.2 m ( Fe(NO 3) m 3 NO 3 ) ( ) = NO 1 Au(NO 3 ) NO 3 ) ( ) = NO 1 Fe(NO 3 ) 3 3 M NO3 = NO NO 3 1 (80.5 m 47.2 m) ( 1000 m ) 1 2 = M NO3 = A sample of 4.95 of solid CH 3COOH is weihed into an Erlenmeyer flask. This sample is titrated with a sodium hydroxide solution, and m of NaOH are required to reach the endpoint. The sodium hydroxide solution is then used to titrate a sample of phosphoric acid of unknown concentration. It requires m of NaOH to react with m of H 3PO 4 solution. (a) Write the balanced equations for the reactions that take place. (b) What is the concentration of the phosphoric acid? CH 3 COOH (s) NaOH (aq) CH 3 COONa (aq) H 2 O (l) M H3 PO 4 = 2.78 H 3 PO 4 (aq) 3 NaOH (aq) Na 3 PO 4 (aq) 3 H 2 O (l) MM CH3 COOH = 2(12.01) 2(16.00) 4(1.008) = n CH3 COOH = 4.95 ( 1 CH 3COOH ) = CH 3 COOH (and NaOH) M NaOH = = m ( 1000 m ) Reaction with H 3 PO 4 : n NaOH = m ( NaOH ) ( 1 H 3PO m 3 NaOH ) = H 3 PO M H3 PO 4 = = m ( 1000 m )

4 7. An 8.65 sample of an unknown roup 2A metal hydroxide is dissolved in 85.0 m of water. An acid-base indicator is added and the resultin solution is titrated with 2.50 M HCl (aq) solution. The indicator chanes color sinalin that the equivalence point has been reached after 56.9 m of the hydrochloric acid solution has been added. (a) What is the ar mass of the metal hydroxide? (b) What is the identity of the metal cation: Ca,Sr, Ba? M(OH) 2 (aq) 2 HCl (aq) MCl 2 (?) 2H 2 O (l) 2.50 HCl n HCl = 56.9 m ( 1000 m ) = HCl MM hydroxide = 122 Cation identity = Sr n M(OH)2 (aq) = HCl ( 1 M(OH) 2 ) = M(OH) 2 HCl 2 MM M(OH) 2 = = MM of M = (2 MM O 2MM H ) = = (Sr2 )

5 8. A solution of 50.0 m of 1.50 M silver nitrate is mixed with a solution of 35.0 m of 1.20 M sodium sulfide. a. Write the balanced ecular and net ionic equations. b. Find the mass of the precipitate formed. Write a table of amounts. The. R. is: A c. Find the concentrations (M s) of all ions at the end of the rxn. x = ANO 3 (aq) Na 2 S (aq) A 2 S (s) 2 NaNO 3 (aq) 2 A (aq) S 2 (aq) A 2 S (s) 50.0 m ( 1.50 ANO m ) = ANO A and NO 3 m ppt () = 9.29 M A = 0 M NO3 = M Na = M S 2 = m ( 1.20 Na 2S ) = Na 1000 m 2 S Na and S A 1 S2 ( 2 A ) = S2 needed (A is the. R. ) 2 2 A (aq) S (aq) A 2 S (s) Initial Chane 2x x x End x x x x = 0 ; x = n A2 S = x = ; MM A (s) 2S = 2( ) = m Mn(OH)2 = ( A 2 S ) = A 2 S M A = 0 (. R. ) ; M S 2 = x V total = = (50.0 m 35.0 m) ( 1000 m ) M NO3 = NO = Expectator ions only et diluted: Na ; M Na = =

6 9. In a double displacement reaction, ml of a M lead (IV) nitrate solution is mixed with m of a M ammonium phosphate solution. a. Write the balanced ecular and net ionic equations. 3 The. R. is: PO 4 b. Find the mass of the precipitate formed. Write a table of amounts. c. Find the concentrations (M s) of all ions at the end of the rxn. x = Pb(NO 3 ) 4 (aq) 4 (NH 4 ) 3 PO 4 (aq) Pb 3 (PO 4 ) 4 (s) 12 NH 4 NO 3 (aq) 3 Pb 4 3 (aq) 4 PO 4 (aq) Pb 3 (PO 4 ) 4 (s) m ppt () = 7.51 [lead (IV)] = [nitrate] = m ( Pb(NO 3) 4 ) = Pb(NO 1000 m 3 ) 4 [ammonium] = [phosphate] = Pb 4 ; NO m ( (NH 4) 3 PO 4 (aq) 1000 m ) = (NH 4 ) 3 PO 4 (aq) NH 4 ; PO Pb 4 ( 4 PO Pb 4 ) = PO 4 3 (needed) ; PO 4 3 is the. R.! 4 3 Pb (aq) 3 4 PO 4 (aq) Pb 3 (PO 4 ) 4 (s) x 4x x x x x x = 0 ; x = MM Pb 3 (PO 4 ) 4 (s) = 3(207.2) 4(30.97) 16(16.00) = / mass Pb 3 (PO 4 ) 4 (s) = ( ) = Pb 1 Pb 3 (PO 4 ) 3 (PO 4 ) 4 (s) 4 (s) [PO 3 4 ] = 0 (. R. ) ; [Pb 4 ( ( )) ] = = 1 (500.0m 200.0m) ( 1000m ) The ions NO 3 and NH 4 are spectator and only et diluted. Their final concentrations are: = [NH 4 ] = NH = ; [NO 3 ] = NO =