Using the above info. Caclulate the stoichiometric concentration of the two impo

Question

Using the above info. Caclulate the stoichiometric concentration of the two important species in the buffer using a NIE and ICE chart.

Also find the concentration of HAc in buffer and Ac- in buffer.

THe following is the graph of the titration:

TITRATION CURVES, AND INDICATORS RS, LAB REPORT WORKSHEETS EXP 5 BUFFERS, Part 2- Preparation of a 1:1 Buffer the space below, write the net ionic equation for the reaction that occurred NaOH was added to acetic acid to prepare the buffer solution. This treated as if it goes to completion, since the value for Keq is a very reaction is large number. Find Keg for yourself below, construct an ICE table. Directly under each species in the 2. In the space NIE calculate the: al "initial" number of moles of the reagents (you know the volumes of both HAc and NaOH, the concentration of NaOH from standardizing the solution and you just calculated the concentration of the stock HAc in Part b) limiting reagent, and therefore the amount of moles that change during the reaction c) final number of moles of each reactant and product, after the reaction has occurred. Divide the final number of moles of each reactant by the total volume of the buffer solution to obtain the stoichiometric concentration of each of the two important species in the buffer. 3. Hint: The two concentrations should be equal or very nearly equal, and will probably lie within 0.04-0.06 M. Volume of Acetic Acid used in the titration Volume of NaOH to reach the equivalence point 16.02 m (from your graph) Volume of NaOH required to make a 1:1 buffer (h of the equivalence point volume) Concentration of "unknown" acetic acid calculated in Part 1: O.lFtMal Calculate the stoichiometric concentrations of the two important species in the buffer: NIE: Stoichiometric concentration of HAc in buffer Stoichiometric concentration of Ac" in buffer

Explanation / Answer

Moles of NaOH used = concentration*volume = 0.1072*0.048 = 0.005 mol

Moles of acetic acid used = concentration*volume = 0.1717*0.060 = 0.010 mol

For ICE table: the reaction is:

NaOH + AcH ----> Ac-Na+ + H2O

This can also be written as:

OH- + AcH ----> Ac- + H2O

ICE table:

Total volume = 60.00+48.00 = 108.00 ml = 0.108 L

Moles of AcH left = 0.005 mol

So, concentration of AcH = moles/volume = 0.005/0.108 = 0.046 M

Concentration of Ac- = 0.005/0.108 = 0.046 M

OH- AcH Ac- Initial 0.005 0.010 0 Change -0.005 -0.005 +0.005 Equilibrium 0 0.005 0.005

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