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ooooo T-Mobile LTE 11:16 PM organic.cc.stonybrook.edu Enter Your Answer: For your pH titration you now want to weigh an amount of your unknown acid based on your answer However, you do not want to waste 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 S13.9 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 mLof NaOH required to reach the endpoint for 513.9 mg of your acid. Enter a numeric a nswer to one decimal place. Do not include units in your answer. 3 For your pH titration you now want to weigh an amount of your unknown acid based on your answer 2. Hi owever, you do not want to waste to question lot of time trying to get the amount exactly. And even if you could, it is not necessary. Using 513.9 mg of your sample, dissolving it in water and titrating with 0.1258 M NaoH produces the pH titration curve shown below Reading from the above curve, what volume of NaOH (mL) was used to reach the equivalence point Depending on your titration technique, it may not be exactly what you calculated Question nswer only, to three significant Enter a numeric a figures. Do not include units. Enter Your Answer: How many mmol of NaOH have been added at the titration equivalence point? Enter a numeric answer to three significant figures. 18%Explanation / Answer
Molarity of NaOH = 0.1258 M
Volume of NaOH added to reach equivalence point = 25 mL = 0.025 L
Moles of naOH added = molarity * volume in L = 0.1258 M * 0.025 L = 3.15*10^-3 moles
= 3.15 mmol (answer)
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