A student weighs 1.700 g of succinic acid and dissolves it in water in a 250.0 m

Question

A student weighs 1.700 g of succinic acid and dissolves it in water in a 250.0 mL. volumetric flask. A 25.00 mL sample of this solution is withdrawn and placed in a 125 mL Erlenmeyer flask, to which 5 drops of the acid-base indicator phenolphthalein is added. This solution is titrated with a sodium hydroxide solution of unknown molarity in a buret. a) How many moles of succinic acid were weighed and dissolved in the volumetric flask? b) How many moles of succinic acid were placed in the Erlenmeyer flask? c) If the unknown sodium hydroxide solution was known to be approximately 0.20 M, at approximately what buret volume would you expect the end point of the titration? Show the calculation, including the use of the appropriate coefficients, as a 2 significant figure estimate: d) If the measured volume at the end point was 13.32 mL of sodium hydroxide added, calculate the actual molarity of the sodium hydroxide solution. You are going to standardize a 0.80% solution of sodium hydroxide solution for which the molarity is approximately 0.20 M. You will have dry succinic acid, a 250 mL volumetric flask, a 25.00 mL pipet, a 50.00 mL buret, a phenolphthalein indicator solution, and other standard laboratory equipment. For each titration you will withdraw a 25.00 mL sample of the succinic acid solution with a volumetric pipet. Therefore, you will deliver exactly one-tenth of your initial mass of succinic acid (to four significant figures) in each titration. In deciding the amount of succinic acid to weigh, assume that you will be using approximately 15 mL of the sodium hydroxide of approximate 0.20 M concentration for each titration, and you need enough succinic acid for ten titrations. Show how to calculate the mass of succinic acid you will need to weigh on the back of this page. (Record the "Planned succinic acid mass" in your lab notebook, and include the reaction equation from above.)

Explanation / Answer

Q2.

m = 1.7 g of acid

V = 250 mL = 0.25 L

sample of V = 25 mL is taken

a)

how many moles of acid

MW of Succinic acid = 118.09 g/mol

so...

mol = mass/MW = 1.7/118.09 = 0.01439 mol of succinic acid in original sample

b)

moles of acid in Erlenmeyer Flask

only 25 mL of 250 mL is kept in flast, so this is 10%

0.01439 *0.10 = 0.001439 mol of succinic acid present

c)

if [NaOH] = 0.20 M, find volume

so..

mol of acid = 2 mol of base

mol of acid = 0.001439 mol of acid

so..

mol of base = 0.001439 *2 = 0.002878 mol of base

M = mol/V

V = mol/M = (0.002878)/0.2 = 0.01439 liters 14.39 mL of base approx

d)

if V = 13.32 mL, find molarity of NaOH

M = mol/V = 0.002878 /(13.32*10^-3)

M = 0.216066 mol per liter of NaOH

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