You will do two different titrations of your unknown acid with NaOH First Indica

Question

You will do two different titrations of your unknown acid with NaOH 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 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. Enter Your Answer: question 2 For your pH titration you now want to weigh an amount of your unknown acid based on your answer to Question 1. However, you do not want to waste a lot of time trying to get the amount exactly. And even if you could, it is not necessary. Let's say you weigh out an amount equal to 513.9 mg (Close Enough But since this amount is different than that you calculated for 25 mL of NaOH, you can no longer expect the equivalence point to be at 25 mL Use the same ratio (mg acid titrated mL of NaoH) from above to now calculate the mL of NaOH required to reach the endpoint for 513.9 mg of your acid. Enter a numeric answer to one decimal place. Do not include units in your answer. Enter Your Answer:

Explanation / Answer

1. If 28.02 ml of NaOH was used to titrate 0.5658 g of unknown acid,

then when 25 ml of NaOH was used,

mass of acid to be used = 25 x 0.5658/28.02 = 0.505 g

that is 505 mg acid consumed/25 ml of NaOH

2. when 0.5139 g of acid is taken,

equivalence volume of NaOH consumed = 25 x 513.9/505 = 25.44 ml

So,

513.9 mg acid consumed/25.44 ml NaOH to reach equivalence point

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