di27 Posted May 15, 2011 Report Share Posted May 15, 2011 Aim: The aim is to determine the concentrations of two acids and consequently the enthalpy of neutralisation for each reaction.Chemicals: NaOH 1M (standardised)HCl 2.0 MCH3COOH 2.0 MSo the formulae would be:NaOH + HCl --> NaCl + H20NaOH + CH3COOH --> NaCH3COO + H2OSo basically, I added 50 cm3 of NaOH to a polystyrene cup and then put the HCl in the burette. I added successive portions of 5cm3 and saw the temperature until I reached 50cm3 of HClDid the same for CH3COOH.Now we have to find the enthalpy of neutralisation valuesQuestions to answerThe enthalpy of neutralisation of a strong acid/base is constant. Why is the value constant?Experimental results for HCL are usually a little less negative than the literature value. Suggest two reasons for this.Suggest a reason why the enthalpies of neutralisation for the reactions involving weak acids and weak bases are always less negative than for strong acids and bases.I think I have to use the formula: Q=mc/_\T but for the m = mass of NaOH + HCl but for the mass of HCl, do I use the volume until it reaches the maximum temperature or do I use 50cm3? Reply Link to post Share on other sites More sharing options...
suggaplum Posted May 16, 2011 Report Share Posted May 16, 2011 Aim: The aim is to determine the concentrations of two acids and consequently the enthalpy of neutralisation for each reaction. Chemicals: NaOH 1M (standardised) HCl 2.0 M CH3COOH 2.0 M So the formulae would be: NaOH + HCl --> NaCl + H20 NaOH + CH3COOH --> NaCH3COO + H2O So basically, I added 50 cm3 of NaOH to a polystyrene cup and then put the HCl in the burette. I added successive portions of 5cm3 and saw the temperature until I reached 50cm3 of HCl Did the same for CH3COOH. Now we have to find the enthalpy of neutralisation values Questions to answer The enthalpy of neutralisation of a strong acid/base is constant. Why is the value constant? Experimental results for HCL are usually a little less negative than the literature value. Suggest two reasons for this. Suggest a reason why the enthalpies of neutralisation for the reactions involving weak acids and weak bases are always less negative than for strong acids and bases. I think I have to use the formula: Q=mc/_\T but for the m = mass of NaOH + HCl but for the mass of HCl, do I use the volume until it reaches the maximum temperature or do I use 50cm3? 1. We make the assumption that strong acids and strong alkalis are fully ionised in solution, and that the ions behave independently of each other. For example, dilute hydrochloric acid contains hydrogen ions and chloride ions in solution. Sodium hydroxide solution consists of sodium ions and hydroxide ions in solution. The equation for any strong acid being neutralised by a strong alkali is essentially just a reaction between hydrogen ions and hydroxide ions to make water. The other ions present (sodium and chloride, for example) are just spectator ions, taking no part in the reaction. The full equation for the reaction between hydrochloric acid and sodium hydroxide solution is: NaOH + HCl --> NaCl + H20 but what is actually happening is: H+ + OH- --> H2O If the reaction is the same in each case of a strong acid and a strong alkali, then the enthalpy changes are similar. Even with 2M of HCl and 1M of NaOH in your experiment, since NaOH becomes the limiting reagent, therefore the enthalpy change is the same. 2. There may be heat loss throughout the process causing the calculated value to be less exothermic. However, if you draw a graph and extrapolate the actual heat when added then it shouldn't be that far away from the literature value. Can't exactly think of another reason now, I'll add another one when it comes to me.. 3. Weak acids give less negative values since they only partially dissociate. For example CH3COOH, only around 4% dissociates into its ions thus the enthalpy of neutralisation is less exothermic and the overall reaction will therefore be more endothermic as more energy is needed to break the bonds. Good luck, hope that helped! 1 Reply Link to post Share on other sites More sharing options...
di27 Posted May 16, 2011 Author Report Share Posted May 16, 2011 (edited) Aim: The aim is to determine the concentrations of two acids and consequently the enthalpy of neutralisation for each reaction. Chemicals: NaOH 1M (standardised) HCl 2.0 M CH3COOH 2.0 M So the formulae would be: NaOH + HCl --> NaCl + H20 NaOH + CH3COOH --> NaCH3COO + H2O So basically, I added 50 cm3 of NaOH to a polystyrene cup and then put the HCl in the burette. I added successive portions of 5cm3 and saw the temperature until I reached 50cm3 of HCl Did the same for CH3COOH. Now we have to find the enthalpy of neutralisation values Questions to answer The enthalpy of neutralisation of a strong acid/base is constant. Why is the value constant? Experimental results for HCL are usually a little less negative than the literature value. Suggest two reasons for this. Suggest a reason why the enthalpies of neutralisation for the reactions involving weak acids and weak bases are always less negative than for strong acids and bases. I think I have to use the formula: Q=mc/_\T but for the m = mass of NaOH + HCl but for the mass of HCl, do I use the volume until it reaches the maximum temperature or do I use 50cm3? 1. We make the assumption that strong acids and strong alkalis are fully ionised in solution, and that the ions behave independently of each other. For example, dilute hydrochloric acid contains hydrogen ions and chloride ions in solution. Sodium hydroxide solution consists of sodium ions and hydroxide ions in solution. The equation for any strong acid being neutralised by a strong alkali is essentially just a reaction between hydrogen ions and hydroxide ions to make water. The other ions present (sodium and chloride, for example) are just spectator ions, taking no part in the reaction. The full equation for the reaction between hydrochloric acid and sodium hydroxide solution is: NaOH + HCl --> NaCl + H20 but what is actually happening is: H+ + OH- --> H2O If the reaction is the same in each case of a strong acid and a strong alkali, then the enthalpy changes are similar. Even with 2M of HCl and 1M of NaOH in your experiment, since NaOH becomes the limiting reagent, therefore the enthalpy change is the same. 2. There may be heat loss throughout the process causing the calculated value to be less exothermic. However, if you draw a graph and extrapolate the actual heat when added then it shouldn't be that far away from the literature value. Can't exactly think of another reason now, I'll add another one when it comes to me.. 3. Weak acids give less negative values since they only partially dissociate. For example CH3COOH, only around 4% dissociates into its ions thus the enthalpy of neutralisation is less exothermic and the overall reaction will therefore be more endothermic as more energy is needed to break the bonds. Good luck, hope that helped! WOW! OMG thank you sooo much!! That really helped and I actually got it! haha also, that means the results have to be exothermic right? But then the results I get aren't negative values...they're positive, unless I just need to add a negative sign. So, Calculations NaOH + HCl --> NaCl + H2O Conc. 1.00 1.67 Vol. 0.05 0.03 n 0.05 0.05 0.05 0.05 So the concentration of HCl used was 1.67 M Q = m x c x ∆T Q = [(50+30) ÷1000] x 4.18 x 8.3 Q = 2.78÷0.05 Q = 55.5 kJmol-1 NaOH + CH3COOH --> NaCH3COO + H2O Conc. 1.00 1.54 Vol. 0.05 0.0325 n 0.05 0.05 0.05 0.05 So the concentration of CH3COOH used was 1.54 M Q = m x c x ∆T Q = [(50+32.5) ÷1000] x 4.18 x 8 Q = 2.76 ÷ 0.05 Q = 55.2 kJmol-1 So do I just add a negative sign to the Q value? Edited May 16, 2011 by di27 Reply Link to post Share on other sites More sharing options...
suggaplum Posted May 16, 2011 Report Share Posted May 16, 2011 Aim: The aim is to determine the concentrations of two acids and consequently the enthalpy of neutralisation for each reaction. Chemicals: NaOH 1M (standardised) HCl 2.0 M CH3COOH 2.0 M So the formulae would be: NaOH + HCl --> NaCl + H20 NaOH + CH3COOH --> NaCH3COO + H2O So basically, I added 50 cm3 of NaOH to a polystyrene cup and then put the HCl in the burette. I added successive portions of 5cm3 and saw the temperature until I reached 50cm3 of HCl Did the same for CH3COOH. Now we have to find the enthalpy of neutralisation values Questions to answer The enthalpy of neutralisation of a strong acid/base is constant. Why is the value constant? Experimental results for HCL are usually a little less negative than the literature value. Suggest two reasons for this. Suggest a reason why the enthalpies of neutralisation for the reactions involving weak acids and weak bases are always less negative than for strong acids and bases. I think I have to use the formula: Q=mc/_\T but for the m = mass of NaOH + HCl but for the mass of HCl, do I use the volume until it reaches the maximum temperature or do I use 50cm3? 1. We make the assumption that strong acids and strong alkalis are fully ionised in solution, and that the ions behave independently of each other. For example, dilute hydrochloric acid contains hydrogen ions and chloride ions in solution. Sodium hydroxide solution consists of sodium ions and hydroxide ions in solution. The equation for any strong acid being neutralised by a strong alkali is essentially just a reaction between hydrogen ions and hydroxide ions to make water. The other ions present (sodium and chloride, for example) are just spectator ions, taking no part in the reaction. The full equation for the reaction between hydrochloric acid and sodium hydroxide solution is: NaOH + HCl --> NaCl + H20 but what is actually happening is: H+ + OH- --> H2O If the reaction is the same in each case of a strong acid and a strong alkali, then the enthalpy changes are similar. Even with 2M of HCl and 1M of NaOH in your experiment, since NaOH becomes the limiting reagent, therefore the enthalpy change is the same. 2. There may be heat loss throughout the process causing the calculated value to be less exothermic. However, if you draw a graph and extrapolate the actual heat when added then it shouldn't be that far away from the literature value. Can't exactly think of another reason now, I'll add another one when it comes to me.. 3. Weak acids give less negative values since they only partially dissociate. For example CH3COOH, only around 4% dissociates into its ions thus the enthalpy of neutralisation is less exothermic and the overall reaction will therefore be more endothermic as more energy is needed to break the bonds. Good luck, hope that helped! WOW! OMG thank you sooo much!! That really helped and I actually got it! haha also, that means the results have to be exothermic right? But then the results I get aren't negative values...they're positive, unless I just need to add a negative sign. So, Calculations NaOH + HCl --> NaCl + H2O Conc. 1.00 1.67 Vol. 0.05 0.03 n 0.05 0.05 0.05 0.05 So the concentration of HCl used was 1.67 M Q = m x c x ∆T Q = [(50+30) ÷1000] x 4.18 x 8.3 Q = 2.78÷0.05 Q = 55.5 kJmol-1 NaOH + CH3COOH --> NaCH3COO + H2O Conc. 1.00 1.54 Vol. 0.05 0.0325 n 0.05 0.05 0.05 0.05 So the concentration of CH3COOH used was 1.54 M Q = m x c x ∆T Q = [(50+32.5) ÷1000] x 4.18 x 8 Q = 2.76 ÷ 0.05 Q = 55.2 kJmol-1 So do I just add a negative sign to the Q value? Since the reaction is exothermic, and the change in temperature calculation is TINITIAL - TFINAL. The temperature of TFINAL should be higher than TINITIAL thus generating a negative value. That is probably the reason why your overall enthalpy change is not negative, but since you didn't give the temperature values, I couldn't be certain. Reply Link to post Share on other sites More sharing options...
di27 Posted May 16, 2011 Author Report Share Posted May 16, 2011 Aim: The aim is to determine the concentrations of two acids and consequently the enthalpy of neutralisation for each reaction. Chemicals: NaOH 1M (standardised) HCl 2.0 M CH3COOH 2.0 M So the formulae would be: NaOH + HCl --> NaCl + H20 NaOH + CH3COOH --> NaCH3COO + H2O So basically, I added 50 cm3 of NaOH to a polystyrene cup and then put the HCl in the burette. I added successive portions of 5cm3 and saw the temperature until I reached 50cm3 of HCl Did the same for CH3COOH. Now we have to find the enthalpy of neutralisation values Questions to answer The enthalpy of neutralisation of a strong acid/base is constant. Why is the value constant? Experimental results for HCL are usually a little less negative than the literature value. Suggest two reasons for this. Suggest a reason why the enthalpies of neutralisation for the reactions involving weak acids and weak bases are always less negative than for strong acids and bases. I think I have to use the formula: Q=mc/_\T but for the m = mass of NaOH + HCl but for the mass of HCl, do I use the volume until it reaches the maximum temperature or do I use 50cm3? 1. We make the assumption that strong acids and strong alkalis are fully ionised in solution, and that the ions behave independently of each other. For example, dilute hydrochloric acid contains hydrogen ions and chloride ions in solution. Sodium hydroxide solution consists of sodium ions and hydroxide ions in solution. The equation for any strong acid being neutralised by a strong alkali is essentially just a reaction between hydrogen ions and hydroxide ions to make water. The other ions present (sodium and chloride, for example) are just spectator ions, taking no part in the reaction. The full equation for the reaction between hydrochloric acid and sodium hydroxide solution is: NaOH + HCl --> NaCl + H20 but what is actually happening is: H+ + OH- --> H2O If the reaction is the same in each case of a strong acid and a strong alkali, then the enthalpy changes are similar. Even with 2M of HCl and 1M of NaOH in your experiment, since NaOH becomes the limiting reagent, therefore the enthalpy change is the same. 2. There may be heat loss throughout the process causing the calculated value to be less exothermic. However, if you draw a graph and extrapolate the actual heat when added then it shouldn't be that far away from the literature value. Can't exactly think of another reason now, I'll add another one when it comes to me.. 3. Weak acids give less negative values since they only partially dissociate. For example CH3COOH, only around 4% dissociates into its ions thus the enthalpy of neutralisation is less exothermic and the overall reaction will therefore be more endothermic as more energy is needed to break the bonds. Good luck, hope that helped! WOW! OMG thank you sooo much!! That really helped and I actually got it! haha also, that means the results have to be exothermic right? But then the results I get aren't negative values...they're positive, unless I just need to add a negative sign. So, Calculations NaOH + HCl --> NaCl + H2O Conc. 1.00 1.67 Vol. 0.05 0.03 n 0.05 0.05 0.05 0.05 So the concentration of HCl used was 1.67 M Q = m x c x ∆T Q = [(50+30) ÷1000] x 4.18 x 8.3 Q = 2.78÷0.05 Q = 55.5 kJmol-1 NaOH + CH3COOH --> NaCH3COO + H2O Conc. 1.00 1.54 Vol. 0.05 0.0325 n 0.05 0.05 0.05 0.05 So the concentration of CH3COOH used was 1.54 M Q = m x c x ∆T Q = [(50+32.5) ÷1000] x 4.18 x 8 Q = 2.76 ÷ 0.05 Q = 55.2 kJmol-1 So do I just add a negative sign to the Q value? Since the reaction is exothermic, and the change in temperature calculation is TINITIAL - TFINAL. The temperature of TFINAL should be higher than TINITIAL thus generating a negative value. That is probably the reason why your overall enthalpy change is not negative, but since you didn't give the temperature values, I couldn't be certain. Okay, wait. Lemme get this straight. The initial temperature for NaOH was 23.7°C And the maximum temperature it reached was 32°C So the ∆T = 23.7-32 = -8.3 Is that right? So the final answer would be negative then. Right! Got it Thanks so much! Reply Link to post Share on other sites More sharing options...
adeljithu Posted October 9, 2013 Report Share Posted October 9, 2013 could u give a sample lab report 1 Reply Link to post Share on other sites More sharing options...
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