First - Indicator Titration using phenolphthalein as a colored indicator Titrati
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Question
First - Indicator Titration using phenolphthalein as a colored indicator Titrating to the equivalence point will give you a ratio of (mg acid titrated/mL of NaOH consumed). Second - pH titration From the above ratio (mg acid/mL NaOH), you can calculate the mass of your unknown acid that will require 25 +/- 3 mL of NaOH to reach the equivalence point. Thus, using this mass of acid in your titration here, you will know to expect the equivalence point to occur at approximately 25 mL of NaOH added. If done correctly, constructing a graph (pH vs vol NaOH added) of your results will give you very accurate values of molar mass and pka In your first titration, using phenolthalein as the indicator, you weight out 565.8 mg of your unknown acid sample. Dissolving this amount in water and titrating with NaOH you find that 28.02 mL are required to reach the equivalence point. What mass (mg) of your sample should you weigh out in order to consume 25 mL of the titrant in your pH titration ? Enter a numeric answer only, do not include units.Explanation / Answer
Titration to equivalance point gives the ratio of
mg of acid / mL NaOH
565.8 mg / 28.02 mL
Hence mass of sample to consume 25 mL of is
25 mL *(565.8 mg/28.02 mL)
= 504.8 mg
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