Mod 6 multiple choice (1 Viewer)
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B and C are acids so produce H+. This H+ reacts with chromate to produce more dichromate. A is a neutral salt so wont produce any H+ or OH-. However D is a basic salt so will produce OH- which reacts with H+ to form water. Since there is an increase in conc of H20 equilbrium shifts to the left (as the h20 reacts with dichromate to form chromate and H+). So D
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So, the goal of this question is to decrease the concentration of dichromate. First, we should identify which one is the dichromate which is clearly Cr2O7 2- as per the name (di- = 2 and chromate = chromium and oxygen compound). As a result, we are trying to shift the equilibrium to the left.
Now we can assess each option to determine which is correct:
A - NaCl does nothing as Na+ and Cl- are mostly spectator ions and do not affect this reaction
B - HCl does affect the equilibrium as it increases the concentration of the H+ ions due to acidic dissociation. By LCP, we clearly know that the system will favour the decrease of H+ ions and shift the equilibrium to the 32.9667156687, not left, making it the wrong option
C - CH3COOH does the same thing as HCl but at a weaker extent but nevertheless doesn't shift the equilibrium to the right. Hence this is the wrong option
D - CH2COOHNa dissociates into CH2COOH- and Na+ (which is a spectator ion). What this does is reduce the concentration of H+ due to the following reaction -
CH2OOH- + H+ <- -> CH3COOH
Hence, to compensate for the loss of H+, by LCP the equilibrium shifts to the left making it the correct option.
Hope this helps.
Now we can assess each option to determine which is correct:
A - NaCl does nothing as Na+ and Cl- are mostly spectator ions and do not affect this reaction
B - HCl does affect the equilibrium as it increases the concentration of the H+ ions due to acidic dissociation. By LCP, we clearly know that the system will favour the decrease of H+ ions and shift the equilibrium to the 32.9667156687, not left, making it the wrong option
C - CH3COOH does the same thing as HCl but at a weaker extent but nevertheless doesn't shift the equilibrium to the right. Hence this is the wrong option
D - CH2COOHNa dissociates into CH2COOH- and Na+ (which is a spectator ion). What this does is reduce the concentration of H+ due to the following reaction -
CH2OOH- + H+ <- -> CH3COOH
Hence, to compensate for the loss of H+, by LCP the equilibrium shifts to the left making it the correct option.
Hope this helps.
As the hydroxide ions react with the hydrogen ions, the concentration of hydrogen ions falls. The system shifts left to minimise the decrease in hydrogen ion concentration by replacing some of the lost hydrogen ions (Le Chatelier's Principle) causing the concentration of dichromate ions to fall.
It is unwise to reason in terms of changes in concentration of the solvent as this is fairly constant. If the effect were due to formation of water rather than loss of hydrogen ions, I can argue (A) adds water and so should also produce a shift left, and yet it doesn't.
Note, this is an example of a flawed question as any significant increase in volume from the addition of any aqueous solution to the system will cause the concentration of all species to fall. The result will be an increase in the value of Q which would consequently cause a shift left. So, without considering the effect of the acidity or basicity of the substance added, the dilution from simply adding an aqueous solution would cause an immediate decrease in dichromate concentration, followed by a shift left and a further decrease in dichromate ion concentration.
The greatest decrease will be caused by (D), assuming the volume of added solution is the same in each case, and the least in (B) where the added HCl will add a shift right to the changes occurring (which will be greater than is caused by the weak acid in (C)).
The greatest decrease will be caused by (D), assuming the volume of added solution is the same in each case, and the least in (B) where the added HCl will add a shift right to the changes occurring (which will be greater than is caused by the weak acid in (C)).