An acid or base sample cannot always be titrated directly, due to producing a ha

Question

An acid or base sample cannot always be titrated directly, due to producing a hard to find equivalence point or due to time sensitive chemicals that will experience alternate reactions (e.g. reacts with air) or physical change (e.g. evaporates) during the time required for the titration process. Step 1: let the unknown acid or base be the limiting reagent in a quick neutralization reaction with standard solution #1, which is a strong base or strong acid that will be in excess. Know the initial amount of standard #1 added precisely (the amount of excess is unknowable at this point). Step 2: in another acid-base neutralization reaction, titrate the solution containing excess of standard #1, using a standard solution #2. Standard solution #2 will be a base if standard #1 is an acid, or an acid if standard #1 is a base. This titration uses standard #2 to determine the amount of excess standard #1 remains from the first reaction. Step 3: calculate the amount of standard #1 consumed in the first step: Moles standard #1 consumed - initial mole standard #1 - excess moles standard #1 Step 4: use stoichiometry to determine the moles of unknown from the moles of standard #1 that were used in the first step. Advantage: Step 2 can be done with standards that are not time sensitive, so one works at a normal pace. Advantage: Step 2 is a regular titration that is optimized for easily finding the equivalence point: a titration using a strong acid and a strong base leads to a predictable equivalence point pH of 7, so we know to a chemical indicator that works for pH = 7. A student pipets 2.000 mL of cloudy ammonia solution into an empty 250.0 mL Erlenmyer flask. 25.00 mL of 0.2000M HCl solution is immediately added. The solution is swirled. The resulting mixture is titrated with a solution of 0.05000 M Ca(OH)_2. An equivalence point is found at 6.83 mL of Ca(OH)_2 solution. Determine the concentration of the original ammonia solution. A. Determine the moles of calcium hydroxide added to reach the equivalence point in the titration. B. Determine the moles of hydrochloric acid were consumed during the titration. C. Determine the original moles of hydrochloric acid that were added to the ammonia solution. D. Determine the moles of hydrochloric acid that reacted with the ammonia. E. Determine the original moles of the ammonia. F. Determine the original molarity of the ammonia solution.

Explanation / Answer

a)

The moles of Ca(OH)2 added to reach equivalence point

Volume of Ca(OH)2 to reach equivalence point = 6.83mL

Molarity of Ca(OH)2 = 0.05000M

Moles of Ca(OH)2 = 6.83 mL x 0.05000 mmol/mL

Moles of Ca(OH)2 = 0.3415 mmol

b)

               2HCl + Ca(OH)2 ------->   CaCl2 + 2H2O

From the above equation it is clear that for every mole of Ca(OH)2   two moles of HCl is consumed

So for 0.3415 mmol of Ca(OH)2, 2 x 0.3415 mmoles of HCl is consumed

Moles of HCl consumed is 0.683 mmoles

c)

original moles of HCl added to ammonia

= 25 mL x 0.2000 mmol / mL

original moles of HCl added to ammonia 5 mmol

d)

Moles of HCl reacted with ammonia

original moles - moles consumed during titration

= (5 - 0.683)mmol

Moles of HCl reacted with ammonia is 4.317

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