I have added some comments to all of them but you need to go through and expound on what I have written.
why would the mixture between 0.1 M acetic acid (conductivity of 7) and 0.1 M ammonia (conductivity of 8) result in a conductivity of 8?
Both acetic acid and aqueous ammonia are weak electrolytes (ionized less than 100%) and they are being replaced with a strong electrolyte (ionized 100%).
Net ionic equation: HC2H3O2 (aq) + NH3 (aq) -> NH4+ (aq) + C2H3O2^- (aq)
Explanation: We see an increase (7 to 8) in conductivity because the products are completely ions like NH4^+ (aq) and C2H3O2^- (aq) in the net ionic equation.
why would the mixture between 0.1 M sulfuric acid (conductivity of 9) and 0.1 M barium hydroxide (conductivity of 7) result in a conductivity of 7?
Net ionic equation: 2H^+ (aq) + SO4^-2 (aq) + Ba^+2 (aq) + OH^-(aq) -> 2H2O (l) + BaSO4 (s)
Explanation: We see a decrease (9 to 7) in conductivity due to the fewer ions started with in reactant compared to the products, which consist of a solid and molecule.
Did you start with fewer ions? You ended up with fewer ions. I THINK you have said it right BUT your explanation is confusing to read(at least to me). I would simply say that BaSO4 is insoluble and H2O is only slightly ionized; therefore, there are fewer ions in the products of the reaction.
why would the mixture between 0.1 M copper (II) sulfate (conductivity of 9) and 0.1 M hydrosulfuric acid (weak conductivity of electricity) result in a conductivity of 9?
Net ionic equation: Cu^+2 (aq) + H2S(g) -> CuS(s) + H^+ (aq)
We see the conductivity remain the same as the copper (II) sulfate has about the same conductivity as the hydronium ions produced in the reaction. H2S is only slightly ionized and has a low conductivity and the CuS is only slightly soluble and contributes a low conductivity.
why would the mixture between 0.1 M copper (II) acetate (conductivity of 9) and 0.1 M hydrosulfuric acid (weak conductivity of electricity) result in a conductivity of 9?
Net ionic equation: Cu^+2 (aq) + 2C2H3O2^- (aq) + H2S (aq) -> CuS (s) + 2HC2H3O2 (aq)
We see the conductivity remain the same as the copper (II) acetate.I don't know how to answer this. Frankly, I would expect the conductivity to be less because you have replaced one strong electrolyte (copper acetate) with CuS(only slightly soluble) and acetic acid (a weak acid that ionizes less than 100%).
I'm asked to write the molecular, ionic, and net ionic equation. Then, I'm supposed to provide an explaination of conductivities observed. Below I have written only the net ionic equation and a brief explanation I think of the first two conductivities observed. However, I did not have any clue on how to explain that last two. Can you check and help me with the explanations? Thanks for the help!
why would the mixture between 0.1 M acetic acid (conductivity of 7) and 0.1 M ammonia (conductivity of 8) result in a conductivity of 8?
Net ionic equation: HC2H3O2 (aq) + NH3 (aq) -> NH4+ (aq) + C2H3O2^- (aq)
Explanation: We see an increase (7 to 8) in conductivity because the products are completely ions like NH4^+ (aq) and C2H3O2^- (aq) in the net ionic equation.
why would the mixture between 0.1 M sulfuric acid (conductivity of 9) and 0.1 M barium hydroxide (conductivity of 7) result in a conductivity of 7?
Net ionic equation: 2H^+ (aq) + SO4^-2 (aq) + Ba^+2 (aq) + OH^-(aq) -> 2H2O (l) + BaSO4 (s)
Explanation: We see a decrease (9 to 7) in conductivity due to the fewer ions started with in reactant compared to the products, which consist of a solid and molecule.
why would the mixture between 0.1 M copper (II) sulfate (conductivity of 9) and 0.1 M hydrosulfuric acid (weak conductivity of electricity) result in a conductivity of 9?
Net ionic equation: Cu^+2 (aq) + H2S(g) -> CuS(s) + H^+ (aq)
We see the conductivity remain the same as the copper (II) sulfate.
why would the mixture between 0.1 M copper (II) acetate (conductivity of 9) and 0.1 M hydrosulfuric acid (weak conductivity of electricity) result in a conductivity of 9?
Net ionic equation: Cu^+2 (aq) + 2C2H3O2^- (aq) + H2S (aq) -> CuS (s) + 2HC2H3O2 (aq)
We see the conductivity remain the same as the copper (II) acetate.
4 answers
I think I have seen these questions before. The net ionic equations look OK. The following are comments on the explanations:
#1) OK
#2) There fewer ions in the products (not reactants) which accounts for the decrease in electrical conductivity.
#3) We start with one ion (Cu++) and we end up with one ion H+. so the conductivity remains the same.
#4) Since the number of ions decreased, the conductivity should have decreased, too. the final mixture may have been contaminated.
#1) OK
#2) There fewer ions in the products (not reactants) which accounts for the decrease in electrical conductivity.
#3) We start with one ion (Cu++) and we end up with one ion H+. so the conductivity remains the same.
#4) Since the number of ions decreased, the conductivity should have decreased, too. the final mixture may have been contaminated.
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this was really helpful.
1 answer