M (base X) = M1 = unknown V (HCl) = V2 = 25.0 mL = 0.0250L M (HCl) = M2 = 0.10M ● M1V1=M2V2 sf=2 M1 (0.0502L Base X) = (0.1M HCl)(0.0250L) M1=0.04980=0.050M Base X => The molarity of Base X based on our procedure is 0.050M VI. Analysis & Discussion Questions: 1. Calculate the volume of 0.125 M NaOH needed to neutralize 25.00 mL of 0.100 M HCl. Include a balanced chemical equation in your answer. NaOH(aq) + HCl(aq) = H 2 O(l) + NaCl(aq) sf=3 M1V1=M2V2 (25.00mL HCl)(0.100M)=V2 (0.125M NaOH) V2 = (2.50mol HCl) /(0.125M NaOH) = 20.0mL => 20.0 mL of 0.125M NaOH is needed to neutralize 25.00mL of 0.100M HCl 2. Discuss the titration process - was your titration successful? What was your percentage error for both unknown solutions? (Obtain the actual concentrations from your teacher) Percentage Error(Actual concentration - Theoretical concentration)/Theoretical concentration)*100 Acid R: ((|0.20M - 0.0500M|)/0.0500M)*100%%=3.0*10^2 % (sf=2) Acid Q: ((|0.014M - 0.012M|)/0.012M)*100% = 16.667=17% (sf=2)
Base X: 0.4%(|0.0498M - 0.0500M|)/0.0500M )*100%=0.400%(sf=3) The titration process overall was not successful because we put both the stock solutions and the standardized solution in the wrong place and therefore, we record all the numbers data incorrectly. The percentage errors of the unknowns titrated are very big, especially for acid R, indicating huge sources of error. 3. A student obtained a wet buret from the cart but failed to rinse it with a small amount of the base before starting a titration. Will more or less titrant (base) be required to neutralize the acid? Explain your answer. If the student failed to rinse it with a small amount base before starting a titration, more titrant will be required to neutralize the acid because water is everywhere in the buret instead of the base. Therefore, when adding the base in, those water and the new base will combine, making it more diluted than before (less concentrated). Moreover, there might be some contaminations in the water that would affect the results. As a result, it will take more titrant to make up that diluted part to neutralize the acid. 4. In standardizing the solution of aqueous sodium hydroxide, a chemist overshoots the end point and adds too much NaOH(aq). Would this error result in a calculated concentration of NaOH that was overestimated or underestimated? Explain your reasoning. A chemist overshoots the end point and adds too much NaOH (aq) will receive an incorrect result of its concentration. The standardized NaOH that he made will have a higher concentration than the one he was supposed to make theoretically and therefore, fail the whole process and overestimate the concentration of NaOH. In two solutions with the same amount of solvent, the one with more NaOH will be more concentrated than the one with correct measured amount of NaOH added. 5. What happens when a student performed the titration without adding an indicator? If you forgot to add the indicator, must you automatically repeat or redo the titration? Explain.
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- Fall '15
- Chemistry, ml, Sodium hydroxide