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titration experiment report

Extracts from this document...

Introduction

Part 1: Finding out unknown concentration of HCl. Equation: + 2HCl -> CaCl2 + H2O +CO2 Data collection: Table 1: Raw data for volume of carbonate used to neutralize the acid. Starting record �0.05 ? Finishing record �0.05 ? Volume of HCL �0.05 ? 1 0.4 26.9 26.5 2 10.0 35.6 25.6 3 9.9 37.6 27.7 The Concentration of is 0.0535 mol/ L The Volume of HCL is 30ml Table 2: the actual values required to find the concentration of HCl with certain amount of Sample number Volume of �0.00105 L Number of moles in �0.000055 mol Number of moles in HCl �0.00011 mol Concentration of HCl �0.00367 L/mol 1 0.0265 0.00142 0.00284 0.09333 2 0.0256 0.00137 0.00274 0.09133 3 0.0277 0.00148 0.00296 0.09867 Average 0.0266 0.00142 0.00285 0.09444 Observation: Acid- pink Base- yellow Neutral- orange At the beginning, there was not any change, but as we continued adding the carbonate drop by drop, the pink started to disappear. The change from pink to orange happened in a sudden. We did our best to see the color change, but to stop it at the appropriate measure was very difficult so that we got one yellowish orange too. The average concentration of the HCl solution is 0.0944 �3.89 % M. ...read more.

Middle

Part 2: > Volume of HCl used to neutralize NaOH= finishing record - starting record > n (HCl) =CV since (=from Part B) > n (NaOH) = n(HCl) > C (NaOH) = n(NaOH) / V (NaOH) V=(litters) n= (mol) C= (mol/L) Example of calculations: * Calculation for uncertainties: Volume of (?) Average= (26.5+ 25.6+ 27.7)/3 =26.6 +) 27.7 - 26.6= 1.1 -) 25.6- 26.6= -1 Uncertainty = sum of the absolute values of + and - /2 |1.1|+|(-1)|/ 2 =1.05 * Volume of neutralizing solution used: Volume of used (?) Finishing volume - starting volume= 26.9 - 0.4 =26.5 * Concentration of the neutralizing agent: Concentration of 12.5 g of and 2.5L of water number of moles in n= m/M (mol) = 12.5/ (40.08+ 12.01 + 16.00x3) =0.125 C= n/V (mol/L) = 0.125 / 2.35 =0.0535 * Calculating number of moles: Number of moles in (mol) 26.5? =0.00265L n=CV =0.0535 x 0.00311 =0.00142 number of moles in HCl (mol) n=2() =2 x 0.00142 =0.00284 * Calculating the concentration of the neutralized agent Concentration of HCl (L/mol) C= n/V = 0.00284. 0.03 =0.09333 * Calculating the percentage of error the final concentration of the solution % of error of average concentration of HCl (% M) ...read more.

Conclusion

The change of color in the solution occurred however we did it with extra attention; we have to change our method a little, in order to avoid pouring too much solution into the beaker. Adding amount: It is similar to the error I just suggested above, but because I had to decide where to stop by my eyes, so that the qualitative measurement was not totally reliable. Using Logger Pro would make it possible quantitative to know exact value of pH, so that we know when the solution was neutralized, because the color of solution is yet hard to identify. It is actually difficult to see if you over-poured or did not pour enough unless you know the exact pH value of the solution. Moreover, I think it is one of the causes of the difference among our and the other two results. Situation: Concentration of the solution was not precise since when we washed the equipments, there was always water remaining in them; it enables the concentration to be lower. We could rinse all equipments we planned to use with the original solution and also after we washed them to use them again. It would prevent lowing the concentration of the solution caused by water. Conclusion From our own experiment, we found that the concentration of NaOH is 0.0987 �1.36% M when the concentration of nutaralizing agent, HCl, 0.0944 M �3.89 %. ?? ?? ?? ?? -1- ...read more.

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