A student was given only one graduated cylinder to use lor this expenment. A (to
ID: 962490 • Letter: A
Question
A student was given only one graduated cylinder to use lor this expenment. A (tor using it to measure 50.0 mL ol the assigned acid, the student failed to rinse or dry the cylinder bolore measuring out the 50.5 ml of the base. Would the calculate d delta be higher, lower, or the same as the literature deltaH*uun? Bne,fy oxplam this difference as a result off using only one graduated cytrvder tor the experiment. Explain how the following changes in the procedure tor this experiment would affect the results. A glass beaker was used instead of a pressed polystyrene cup. A pressed polystyrene top was used to cover the pofystyrene cup after the acid and base solutions had been mixed. The accepted delta H robrorm: aod (H r) reacting wIth NaOH solution and of HN03 reacts® with potassium hydroxde (KOH) solution are identical. wnte net tonic equatons to show what aqueous HBr and HN03 have in common. State what aqueous NaOH and KOH have m common. Explain why you would expect that delta for HBr reactng with NaOH solution and delta for HNO_3 reactng with KOH solution would be identical. Went appropriate equabons to support this explanation.Explanation / Answer
1.Some base is neutralised in the cylinder and the corresponding heat contribution is lost.
The measured volume of base was also diminished by the remaining acid solution.Both mistakes introduce negative errors in the measured H value (it will be lower).
2. 1.During the experiment some heat is transferred to the calorimeter and to the environment and is lost for further calculation.
A glass beaker has a higher mass (thus a higher heat capacity of the calorimeter) and higher thermal losses. The measured H value will be lower with a glass beaker.
2.2. It was a good precaution, preventing heat loss to the environment (preventing a negative error).
3. 1.Both are strong acids (completely dissociated in water): the common ion relevant for neutralization is H3O+.
HBr + H2O = H3O+ + Br-
HNO3 + H2O = H3O+ + NO3-
3.2. Both are strong electrolytes (completely dissociated in water): the common ion relevant for neutralization is HO-.
3.3. In both cases the net reaction was
H3O+ + HO- = 2H2O
The same situation will be be for any pair of strong acid + strong base.
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