Acid Base Reactions Help (1 Viewer)

[csi]

1. Given:
OCl- + HCl <-> HOCl + OH-
Which of the below will increase the concentration of HOCl?
A. Adding H2O
B. Adding HCl
C. Adding KCl
D. Adding KOH

2. Describe two properties used to distinguish between a strong monoprotic acid and a weak monoprotic acid of the same concentration (1.00 mol L^-1).

3. A solution of Hal has a pH of 2, what amount (moles) of H+ ions would be present in 500mL of this solution?

4. The molarity of concentrated sulfuric acid is 18 mol L^-1. What volume of concentrated sulphuric acid is required to prepare 1L of 2 mol L^-1 H2SO4 solution?

Thanks

 

[Sp3ctre]

I'll be honest, my chemistry is quite rusty but I'll give it a go, so take this with a grain of salt (someone please correct me if any of my answers are wrong)

1. You want to increase [HOCl] by favouring the forward reaction and shifting equilibrium to the right. Out of all the options, HCl is the only compound which is a reactant in your equation. By adding HCl, the forward reaction will be favoured, which increases concentration of your products. So B should be right.

2. The more acidic a solution is, the lower its pH is. Strong acids completely dissociate into H+ ions in water and thus will have much lower pH than weak acids, which only partially dissociate into H+ ions. Strong acids are better electrical conductors than weak acids since they fully ionise and produce more mobile charges in solution.

3. Idk what Hal is, but seems like you just need to work out the mole ratio between Hal and H+ then just use the pH formula.

4. Also not 100% about this one but it looks like you can use n = cv twice, firstly to find the moles of the 2 mol/L soln then secondly to find the volume of the 18 mol/L sol.

 

[csi]

I'll be honest, my chemistry is quite rusty but I'll give it a go, so take this with a grain of salt (someone please correct me if any of my answers are wrong)

1. You want to increase [HOCl] by favouring the forward reaction and shifting equilibrium to the right. Out of all the options, HCl is the only compound which is a reactant in your equation. By adding HCl, the forward reaction will be favoured, which increases concentration of your products. So B should be right.

2. The more acidic a solution is, the lower its pH is. Strong acids completely dissociate into H+ ions in water and thus will have much lower pH than weak acids, which only partially dissociate into H+ ions. Strong acids are better electrical conductors than weak acids since they fully ionise and produce more mobile charges in solution.

3. Idk what Hal is, but seems like you just need to work out the mole ratio between Hal and H+ then just use the pH formula.

4. Also not 100% about this one but it looks like you can use n = cv twice, firstly to find the moles of the 2 mol/L soln then secondly to find the volume of the 18 mol/L sol.

Any help is appreciated, thanks

 

[Eagle Mum]

I agree with Sp3ctre’s responses to Q1 & 2.

For Q3, pH is the negative log of the concentration of hydrogen ions, so for any solution of pH of 2:
-log10[H+] = 2
[H+] = 10^-2 mol/L (or 0.01 mol/L)
Therefore, in 500 mLs, there is 0.01/2 = 0.005 moles

For Q4, the diluent isn’t specified so I assume it is meant to be water (BTW, top safety tip: NEVER pour water into concentrated acid, always pour acid into water). A 1 in 9 dilution is required to reduce a concentration of 18 M to 2 M, so to produce a final volume of 1000 mL, the required volume of the original concentrated solution is 1000 / 9 which is 111.1 mL (the quotient is not a whole integer, but 111 mL would be more than sufficient for the required accuracy implied by the 1 sig fig in the question).

 

[csi]

I agree with Sp3ctre’s responses to Q1 & 2.

For Q3, pH is the negative log of the concentration of hydrogen ions, so for any solution of pH of 2:
-log10[H+] = 2
[H+] = 10^-2 mol/L (or 0.01 mol/L)
Therefore, in 500 mLs, there is 0.01/2 = 0.005 moles

For Q4, the diluent isn’t specified so I assume it is meant to be water (BTW, top safety tip: NEVER pour water into concentrated acid, always pour acid into water). A 1 in 9 dilution is required to reduce a concentration of 18 M to 2 M, so to produce a final volume of 1000 mL, the required volume of the original concentrated solution is 1000 / 9 which is 111.1 mL (the quotient is not a whole integer, but 111 mL would be more than sufficient for the required accuracy implied by the 1 sig fig in the question).

tyty

 

[CM_Tutor]

1. Given:
OCl- + HCl <-> HOCl + OH-
Which of the below will increase the concentration of HOCl?
A. Adding H2O
B. Adding HCl
C. Adding KCl
D. Adding KOH

2. Describe two properties used to distinguish between a strong monoprotic acid and a weak monoprotic acid of the same concentration (1.00 mol L^-1).

3. A solution of Hal has a pH of 2, what amount (moles) of H+ ions would be present in 500mL of this solution?

4. The molarity of concentrated sulfuric acid is 18 mol L^-1. What volume of concentrated sulphuric acid is required to prepare 1L of 2 mol L^-1 H2SO4 solution?

Thanks

There is something wrong in question 1 as the equation is not balanced.

With q2, watch out on "describe"-ing two properties. The obvious ones to use are indeed pH (lower for the strong acid) and conductivity.

With q3, I suspect HaI is meant to indicate a possible polyprotic acid, like saying H_nX. In any case, @Eagle Mum is correct that it doesn't matter for finding n(H⁺) which depends only on pH and the volume of the solution.

q4 is a 1-in-9 dilution so 111.1 mL sulfuric acid made up to 1000 mL with water.

 

[Eagle Mum]

CM_Tutor is right about Q1 being unbalanced. HCl is also a strong acid.

I would guess the equation they might have intended would be:
OCl- + H2O <-> HOCl + OH-

If that were the case, by Sp3ctre’s correct method of reasoning, the answer would actually be A.

Incorrect questions are usually avoided in exams by rigorous quality checks; if they slip through, the questions are usually discounted, but if I saw this as a question in an exam, I would probably not second guess the correct equation, but answer as Sp3ctre did because in all likelihood that is the reasoning the majority of candidates are likely to apply.

 

[CM_Tutor]

CM_Tutor is right about Q1 being unbalanced. HCl is also a strong acid.

I would guess the equation they might have intended would be:
OCl- + H2O <-> HOCl + OH-

If that were the case, by Sp3ctre’s correct method of reasoning, the answer would actually be A.

Incorrect questions are usually avoided in exams by rigorous quality checks; if they slip through, the questions are usually discounted, but if I saw this as a question in an exam, I would probably not second guess the correct equation, but answer as Sp3ctre did because in all likelihood that is the reasoning the majority of candidates are likely to apply.

I agree with Eagle Mum about the equation that was likely intended, but I don't agree about the answer.

Adding water will (theoretically) lead to a shift right (applying Le Chatelier's Principle), but water being the solvent and effectively constant in concentration means adding it has little practical effect. Further, the addition of water would dilute all of the species, decreasing the concentration of both reactants and products.

I believe the best answer is (B) and I would expect that to be the only accepted answer if the equation suggested above was given. The reason is that adding hydrochloric acid would neutralise some of the hydroxide ion, decreasing the concentration of a product and thus causing a shift to the right by applying Le Chatelier's Principle.

The usual approach to MCQ is that the best answer should be given and (B) is unambiguously going to cause an increase in the concentration of HOCl. If water is added and causes a shift to the right, the overall effect could still be a decrease in the concentration of HOCl due to the dilution. That is, we could have:

• [HOCl] is x mol/L
• Water is added, causing [HOCl] to instantly fall to (say) 0.8x mol/L
• The shift to the right in line with Le Chatelier's Principle causes more HOCl to be produced
• Hence, [HOCl] rises back to 0.95x mol/L

I am making these numbers up but they illustrate that the concentration could fall (overall) while rising due to the shift under a scenario from answer (A).


PS: What made me notice the unbalanced equation was considering answer (C). If the equation did include HCl which, being a strong acid, would be ionised to H⁺ and Cl^-, then adding KCl (and hence a source of chloride ions) could cause a shift the the right. So...

• (D) would cause a shift left by adding hydroxide ions, a product
• (C) would have no effect if the equation has no potassium ions or chloride ions, or a shift right with the unbalanced equation given.
• (B) would cause a shift right in the equation as written by adding a reactant and consuming a product (which illustrates that the equation given is strange), and would cause a shift right for the (likely) intended equation by removal of hydroxide ions, a product
• (A) is meant to be treated as having no effect, being a solvent in a case where it does not appear in the equation (Suppose enough water is added to halve the concentration of every species in the equation given... the value of Q would not change as [C][D] / [A][] would become ([C] / 2)([D] / 2) / ([A] / 2)([] / 2), which has the same value.) If water is a reactant, as in the (likely) intended equation, Le Chatelier's Principle says shift right but the disturbance has changed the concentration of all of the species and may cause all species concentrations to fall overall.

 

[Eagle Mum]

I agree with Eagle Mum about the equation that was likely intended, but I don't agree about the answer.

Adding water will (theoretically) lead to a shift right (applying Le Chatelier's Principle), but water being the solvent and effectively constant in concentration means adding it has little practical effect. Further, the addition of water would dilute all of the species, decreasing the concentration of both reactants and products.

I believe the best answer is (B) and I would expect that to be the only accepted answer if the equation suggested above was given. The reason is that adding hydrochloric acid would neutralise some of the hydroxide ion, decreasing the concentration of a product and thus causing a shift to the right by applying Le Chatelier's Principle.

The usual approach to MCQ is that the best answer should be given and (B) is unambiguously going to cause an increase in the concentration of HOCl. If water is added and causes a shift to the right, the overall effect could still be a decrease in the concentration of HOCl due to the dilution. That is, we could have:

• [HOCl] is x mol/L
• Water is added, causing [HOCl] to instantly fall to (say) 0.8x mol/L
• The shift to the right in line with Le Chatelier's Principle causes more HOCl to be produced
• Hence, [HOCl] rises back to 0.95x mol/L

I am making these numbers up but they illustrate that the concentration could fall (overall) while rising due to the shift under a scenario from answer (A).


PS: What made me notice the unbalanced equation was considering answer (C). If the equation did include HCl which, being a strong acid, would be ionised to H⁺ and Cl^-, then adding KCl (and hence a source of chloride ions) could cause a shift the the right. So...

• (D) would cause a shift left by adding hydroxide ions, a product
• (C) would have no effect if the equation has no potassium ions or chloride ions, or a shift right with the unbalanced equation given.
• (B) would cause a shift right in the equation as written by adding a reactant and consuming a product (which illustrates that the equation given is strange), and would cause a shift right for the (likely) intended equation by removal of hydroxide ions, a product
• (A) is meant to be treated as having no effect, being a solvent in a case where it does not appear in the equation (Suppose enough water is added to halve the concentration of every species in the equation given... the value of Q would not change as [C][D] / [A][] would become ([C] / 2)([D] / 2) / ([A] / 2)([] / 2), which has the same value.) If water is a reactant, as in the (likely) intended equation, Le Chatelier's Principle says shift right but the disturbance has changed the concentration of all of the species and may cause all species concentrations to fall overall.

Referring to the equation: OCl- + H2O <-> HOCl + OH-
What if all the species were in a solution with a nonaqueous aprotic polar solvent where H20 was initially present in a very small concentration (ie. a solute which is not in excess)?

 

[CM_Tutor]

Referring to the equation: OCl- + H2O <-> HOCl + OH-
What if all the species were in a solution with a nonaqueous aprotic polar solvent where H20 was initially present in a very small concentration (ie. not in excess)?

Then the effect of an addition of water would be the same as the addition of (say) NaOCl, so long as it dissolved in the solvent, would have as an addition of hypochlorite ions.

 

[Eagle Mum]

Then the effect of an addition of water would be the same as the addition of (say) NaOCl, so long as it dissolved in the solvent, would have as an addition of hypochlorite ions.

In other words, favour the reaction to the right.

I guess your point upthread emphasises the importance of the subscripts in the description of each species.

 

[CM_Tutor]

In other words, favour the reaction to the right.

I guess your point upthread emphasises the importance of the subscripts in the description of each species.

Yes, in the situation that you describe, the reaction would properly be written

OCl^-_(sol) + H₂O_(sol) <------> HOCl_(sol) + OH^-_(sol)

For anyone who has not seen this before, "sol" is the state used for dissolved in a solvent other than water.