concentration of acetic acid of vinegar‚ in terms of (m/m). Introduction Neutralization is a reaction of an acid with a bases to produce a salt and water. For example‚ the reaction of the strong acid HCl with the strong base NaOH produces the salt sodium chloride and water HCl(aq) NaOH(aq) ( NaCl(aq) H2O(l) Since the acid‚ base‚ and salt in this reaction are all strong electrolytes‚ the net ionic equation for this reaction is H(aq) OH-(aq) ( H2O(l) This same result occurs in the reaction of strong
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Experiment No. ___________________ Date ___________________ NEUTRALIZATION TITRATIONS INTRODUCTION The neutralization of hydronium or hydroxide ion to form water is widely used as the basis for volumetric determinations of acids‚ bases and salts of weak acids. The reaction is characterized by a rapid change in pH near the equivalence point‚ a change that is readily detected by an acid-base indicator or that can be followed electrically by use of a pH meter. Neutralization titrations
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In the first week of the experiment‚ the goal was to find the moles of NaOH‚ as well as a 0.1 molarity‚ while in the second week to goal was to determine the percent KPH in the sample. The first week titrations were successful and had very similar amounts of pink‚ which shows the precision of the results. The best trials were in the sample trial and the second and third trials. The average molarity calculated for the first week was 0.1017 M. This very close to the 0.1 M that was supposed to be made
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1957‚ 340-343). Neutralization reactions in experiments: NaOH(aq) + KHP(aq) Na+ + KP- +H2O NaOH(aq) + CH3COOH CH3COO- + Na+ +H2O Infromation - Some indicators (including Phenophthalein) Indicator | pH at which colour changes | Colour at lower pH | Colour at higher pH | Phenolphthalein | 9 | Colourless | Red | Litmus | 7 | Red | Blue | Alizarin Yellow | 11 | Yellow | Red | Methyl Red | 5 | Red | Yellow | - KHP This is a large molecule
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phthalate (KHP‚ formula weight 204.2 g/mol). From the following data‚ calculate the molarity of the NaOH solution: mass of KHP 1.404 g; buret reading before titration 0.13 mL; buret reading after titration 38.57 mL. Answer in units of M 002 10.0 points What is the main objective of Part 1 of the experimental procedure? What is the purpose of using KHP in this part of the experiment? 1. The main purpose of Part 1 is to gain practice doing titrations before we titrate our unknown acid in Part 2. KHP is used
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phthalate and two unknown acids were titrated. We determined the molar mass of the Potassium hydrogen phthalate‚ for the unknown acids we calculated the molar mass and the Ka values. We used NaOH as the known base for titrating in all three of the titrations. Our ka value based on our titration curve of KHP was 6.9 (the equivalence point where the pH equals the pKa) and our unknown acids were determined to be Hexanoic Acid (with an actual molar mass of 116.6g) with a determined mass of 116.1g. The
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chemical solutions; e) to help the student improve his/her lab technique. Theory: Titration was used to study acid-base neutralization reaction quantitatively. In acid-base titration experiment‚ a solution of accurately KHP concentration was added gradually to another solution of NaOH concentration until the chemical reaction between the two solutions were completed. The equivalence point was the point at which the acid was completely reacted with or neutralized by the base. The point was signaled by
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a strong base. The titrant reacts with and consumes the acid via a neutralization reaction. The point at which stoichiometric amounts of the acid and base have combined is the equivalence point. An example of this is shown in the equation: HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l). The number of moles is given by knowing the exact concentration and volume added of the titrant. The latter‚ in turn‚ is related by stoichiometry to the number of moles of acid initially present in the unknown. To detect the equivalence
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Methods/Procedure: First begin by mixing up and standardizing a 500mL solution of NaOH to titrate. For each of the six bottles‚ measure the directed amounts of ester‚ water‚ alcohol‚ and HCl. The bottles of different solutions will be left to come to equilibrium for two weeks. Once the NaOH is standardized‚ the solutions in the bottles have come to equilibrium‚ and a molarity is calculated‚ use the molarity of NaOH to discover how many mols were used to neutralize the solutions in each bottle. Once
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Calculations and Discussions Preparation of 250-mL 0.1 M NaOH Solution: Wt. of NaOH= (vol. of Sol’n) (M of Sol’n) (MW of NaOH) = (250 mL) (0.1 M NaoH) (40.0g/mol NaOH) Wt. of NaOH= 1.00 g • One gram of NaOH pellets was weighed and dissolved in distilled water. The solution was diluted to 250 mL. Table 1.Weighing of KHP (weighing by difference) |Replicate |Wt. of container -sample‚ g |Wt. of KHP | |
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