Experiment #4: Acid-Base Eqanm Goal #2 Prier to Class Calculations: Calculate th

Question

Experiment #4: Acid-Base Eqanm Goal #2 Prier to Class Calculations: Calculate the volume and mass required to make 100 mL of 0.10 M solutions of your assigned acid and conjugate base-see concentrations and molar masses. page 4 for ldentily of Assigned AcialBase Conugate Pair: or ptioie otiol Saduin Ppamaa Concentration of Stock Solution of Acid (or Base in case of Ammonia Actual quantity of acid used Describe the procedure you used for making the solutions for both the acid and the base. Include relevant experimental details like the glassware used. Show the calculation for determining the actual concentrations of solution made with the solid. Acid Base Actual Concentration Measured pH 10

Explanation / Answer

To prepare 100ml of 0.1M propanoic acid/sodium propanoate buffer,

Total mol in solution=0.1mol/L*0.1L=0.01mol

take 5ml propanoic acid(1.0M) in 100 ml volumetric flask.

mol of propanoic acid added=1.0mol/L*0.005L=0.005mol

mol of sodium propanoate to be added=total mol-mol of propanoic acid=0.01mol-0.005mol=0.005mol

mass of sodium propanoate to be added=0.005mol *MW=0.005mol*(96.06g/mol)=0.480g

Now add 0.480g to the above volumetric flask and dilute it upto themark using distilled water.The solution is 0.1M /100ml of propanoic acid/sodium propanoate buffer

propanoic Acid=0.005mol/0.1L=0.0005M

conjugate base (sodium propanoate) concentration=0.005mol/0.1L=0.0005M

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