2.The hydrolysis of an ester is followed by acid-base titration. We take a sample of 30.0 cm3 from the reaction mixture which was originally a 0.02 M NaOH solution. We put the sample 50.0 cm3 0.02 M HCl solution and for the back titration of the acid we needed 25.3 cm3 0.02 M NaOH solution. Calculate the concentration of the ester which reacted.

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Jean KeelingLv2

30 Sep 2019

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In an acid-base titration, 24.88 mL of an NaOH solution are needed to neutralize 26.43 mL of a 0.1049 M HCl solution, To find the molarity of the NaOH solution, we can use the following procedure: a First note the value of MH- in the HCl solution. b. Find MOH- in the NaOH solution. c. Obtain MNaOH from MOH-.

A 0.1270 g sample of solid acid is titrated with 28.21 mL of 0.1095 M NaOH. This overshot the endpoint and 2.51 mL of 0.1064 M HCl was used in the back titration. (Back Titration is when you add excess NaOH and have to use HCl to neutralize the excess) Total moles of NaOH = (moles H+ from HCl) + (moles H+ from sample)

a. Calculate the total moles of NaOH added

b. Calculate the total moles of HCl added

c. Calculate the moles NaOH that only reacted with the solid acid (This also equals the moles of H+ from the solid acid).

d. Calculate the equivalent mass of the acid (grams acid per mole H+).

e. If this is a diprotic acid, what is the molar mass?

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aquamarineprairie-dog413

Please provide the steps for the solution. I am particularly confused with what I'm supposed to subtract the total moles of HCl with. I think the concentration of the solution can be determined via M1V1=M2V2 but not sure