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TITRATION OF A WEAK ACID WITH A STRONG BASE - From the experiment I have calculated that the molar enthalpy of neutralization for sodium hydroxide wiht sulphuric acid is -(54 3.2 kJ/mol).

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Introduction

LAB #7:TITRATION OF A WEAK ACID WITH A STRONG BASE Submitted by Tim Kwok Chemistry 20 IB Presented to Ms. Gierach December 11, 2009 Design Problem: Refer to teacher sheet. Prediction: Materials: * 10mL pipette * a burette * 150mL beakers * pH meter * a retort stand * sodium hydroxide acid solution * weak acid solution Procedure: Refer to teacher sheet. Variables: Data Collection and Processing Quantitative Data: Pipette (mL) ?0.04mL : TITRATION OF WEAK ACID WITH STRONG BASE POTENTIOMETIRC TRIAL Burette Reading (mL) ?0.05mL pH ?0.01 Burette Reading (mL) ?0.05mL pH ?0.01 12.50 2.95 18.55 10.88 16.80 5.18 18.60 11.09 17.00 5.30 18.65 11.20 17.20 5.37 18.70 11.30 17.25 5.40 18.75 11.36 17.50 5.48 18.85 11.45 17.60 5.54 18.90 11.50 17.70 5.64 18.95 11.52 17.80 5.67 19.00 11.56 17.90 5.74 19.05 11.59 18.00 5.84 19.10 11.61 18.05 5.88 19.15 11.63 18.10 5.97 19.30 11.68 18.20 6.11 19.55 11.74 18.25 6.20 19.95 11.77 ...read more.

Middle

= 8.00?0.4�C (0.050L ? 0.0002L NaOH(aq)) * 1mol NaOH(aq) * ?rHm, NaOH = (0.080 ? 0.0004kg H2O(l)) * 4.19 kJ * (8.00? 0.4�C) 1L NaOH(aq) kg * �C (0.050L ? 0.4% NaOH(aq)) * 1mol NaOH(aq) * ?rHm, NaOH = (0.080 ? 0.50% kg H2O(l)) * 4.19 kJ * (8.00? 5.0%�C) 1L NaOH(aq) kg * �C (0.050 ? 0.4% mol NaOH(aq)) * ?rHm, NaOH = (2.6816... ?5.5% kJ) ?rHm, NH4Cl = (2.6816... ?5.5% kJ) (0.050 ? 0.4% mol NaOH(aq)) ?SolHm, NaOH = 53.632... ?5.9% kJ/mol ?SolHm, NaOH = 53.632... ?3.164288... kJ/mol ?SolHm, NaOH = 54 ?3.2 kJ/mol Temperature increases = exothermic reaction = -(KJ/mol) so: ?SolHm, NaOH = -(54 ?3.2 kJ/mol) Percent Error: ?(?57 kJ/mol -54 kJ/mol?) / 57 kJ/mol?*100% = 5.236...% ?(?57 kJ/mol -54 kJ/mol?) / 57 kJ/mol?*100% = 5.3% Conclusion and Evaluation Conclusion: From the experiment I have calculated that the molar enthalpy of neutralization for sodium hydroxide wiht sulphuric acid is -(54 ?3.2 kJ/mol). ...read more.

Conclusion

In actuality the product of the neutralization has it's own specific heat capacity, this means that the calculations are off by a bit since the actual specific heat capacity was not used. Improvements: In order to close the system off even more, you can tape all the cracks that are visible on the calorimeter so that less energy is lost between the system and the environment. To fix the problem of thermal energy being lost by calorimeter material, just calculate the quantity of thermal energy each material loses and subtract it from you final temperature to compensate for the specific heat capacity of the other materials. This gives you the actual Q of the product, and since Q = mc?t and n?rHm = mc?t, n?rHm = Q. The last improvement that can be made to the experiment would be to use the actual specific heat capacity of the product rather than just 4.19 J/g * �C. All these improvements will help bring the calculated value closer to the theoretical value minimizing the percent error. ...read more.

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