LAB (QUIZ 1): Internet Questions Flashcards
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Suppose an unknown metal sulfate is found to be 72.07% SO4. Assuming the charge on the metal cation is +3, determine the identity of the cation.
Iron (Fe)
Fe 3+
The unknown metal sulfates are hygroscopic and will absorb water from air. The unknowns must thus be kept in desiccators to remove any absorbed water. How would your results be affected if your unknown sample was not desiccated? Would this error cause your calculation of the mass percent of sulfate in the unknown to be too high or too low? Explain.
If the unknown metal sulfates are not desiccated and absorb water from the air, it would cause an error in the determination of the mass percent of sulfate in the unknown. The error would make the calculated mass percent of sulfate appear higher than it actually is. This is because the absorbed water would add to the mass of the unknown sample, increasing its total weight, and sulfate content would be calculated based on this higher weight.
Undesicated sample absorbs water, falsely inflating sulfate content.
In this experiment you used an excess of the BaCl2 solution. How would your results be affected if you did not use an excess of the BaCl2 solution? Would this error cause your calculation of the mass percent of sulfate in the unknown to be too high or too low? Explain.
If an excess of the solution is not used, it would result in an incomplete reaction between the unknown metal sulfate and the solution. This would cause an error in the determination of the mass percent of sulfate in the unknown, making it appear lower than it actually is. This is because only a fraction of the sulfate ions would react with the limited solution, leading to an underestimation of the sulfate content in the unknown sample. Using an excess of the solution ensures that all available sulfate ions react completely, providing accurate results.
Insufficient solution yields lower sulfate result; excess ensures complete reaction.
In the last step of the procedure, you vigorously heated the BaSO4 precipitate wrapped in filter paper in a crucible. How would your results be affected if tiny pieces of the filter paper still remained mixed in with the BaSO4 after heating? Would this error cause your calculation of the mass percent of sulfate in the unknown to be too high or too low? Explain.
If tiny pieces of filter paper remained mixed with the BaSO₄ after heating, the measured mass of the BaSO₄ and filter paper mixture would be higher than the actual mass of just the BaSO₄. Since the mass of the precipitate is in the numerator of the mass percent calculation, this inflated mass would lead to a higher calculated mass percent of sulfate in the unknown.
** The filter paper adds extra mass that isn’t sulfate, thus artificially increasing the calculated percentage.**
What was the purpose of the phenolphthalein indicator in this experiment?
Phenolphthalein is a colourless, weak acid that is widely used as an indicator in titration experiments to indicate the endpoint of the titration. The endpoint is indicated by the formation of a pink colour since this compound dissociates to form pink anions when dissolved in water.
Suppose you added 40 mL of water to your vinegar sample instead of 20 mL. Would the titration have required more, less or the same amount of NaOH(aq) for a complete reaction? Explain.
Titration of vinegar (acetic acid) requires a base such as NaOH (sodium hydroxide). The reaction would be
CH3COOH + NaOH ==> CH3COONa + H2O
From the balanced equation, you can see that it takes 1 MOLE of NaOH to neutralize 1 MOLE of CH3COOH. The volume of CH3COOH is not relevant. So, whether you have 20 ml or 10 ml, as long as the moles are the same (you didn’t add or remove any CH3COOH), the volume of NaOH will be the same.
we titrate moles, not concentration
Consider a 0.586 M aqueous solution of barium hydroxide, Ba(OH)2 (aq).
* How many grams of Ba(OH)2 are dissolved in 0.191 dL of 0.586 M Ba(OH)2 (aq)?
* How many individual hydroxide ions (OH) are found in 13.4 mL of 0.586 M Ba(OH)2 (aq)?
* What volume (in L) of 0.586 M Ba(OH)2 (aq) contains 0.466 ounces of Ba(OH)2 dissolved in it
(a) 1. 918 g
(b) 9.45 x 10²¹
(c) 0.131 L or 131 mL
If 16.0 mL of water are added to 31.5 mL of 0.586 M Ba(OH)2(aq), what is the new solution molarity?
0.389 M
Suppose you had titrated your vinegar sample with barium hydroxide instead of sodium hydroxide:
Ba(OH)2(aq)+2HC2H3O2(aq)⟶Ba(C2H3O2)2(aq)+2H2O(l)
What volume (in mL) of 0.586 M Ba(OH)2(aq) must be added to a 5.00 mL sample of vinegar to reach the equivalence point? Use your average vinegar molarity (see page 1) in this calculation.
0.000426 L or 0.426 mL.
