ACTIVITY: SOLUTION CONCENTRATION VS. CONDUCTIVITY Here is your goal for this lesson:

Graph experimental data and interpret results for peer review

A chemistry student carried out an experiment with a conducting apparatus (ammeter) similar to the one below. This ammeter measures in milliamperes (mA). The following data was taken.

Solution Reading
0.1 M H2SO4 150 mA
0.1 M Ba(OH)2 150 mA
To 30 mL of the Ba(OH)2 solution, 10 mL portions of H2SO4 were added until a total of 50 mL of H2SO4 were used. The following results were recorded.

DATA TABLE

Total H2SO4 Added Meter Reading Observations
0 mL 150 mA Ba(OH)2 and H2SO4 clear, colorless
10 mL 65 mA milky white precipitate forms
20 mL 31 mA more precipitate forms
30 mL 0 mA precipitate heavy and settles
40 mL 29 mA no added precipitate seen to form
50 mL 62 mA no change seen

Questions

1. Did you plot the data?
yes
no

2. Did you label your graph axes?
no
yes

3. Did you give your graph a title?
no
yes

4. Does the Ba(OH)2 solution contain ions?
yes
no

5. Does the H2SO4 solution contain ions?
yes
no

6. Explain the data.
Is there any evidence that a reaction has occurred?

7. Does the conductivity increase or decrease?

8. Does the number of ions in solution increase, decrease, or remain constant?

9. What is the indicator of the number of ions in solution?

10. How does this evidence indicate that the reaction has occurred between ions?

11. The Ba(OH)2 dissociates as Ba+2 + 2 OH-. H2SO4 dissociates as 2 H+ + SO4-2.
Write a balanced equation for this reaction.

12. When the conductivity is at a minimum, what must be true about the amount of Ba(OH)2?

13. Why does it not conduct at this low point?

14. Why does it conduct more before and after this minimum point?

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is  2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry  problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that     10 dm = 1 m  ⇔ 10³ dm³ = 1 m³then we use the fact that     10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting:     now we have to find the moles of hydrochloric  acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by    molar mass = (mass ca) + (mass c) + 3(mass o)    = 40 + 12 + 3(16)    = 100 g/molconvert the given 5 g to moles:     mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis    mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃)    = 0.05 mol hcl * 2    = 0.1 mol hcl  we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get     volume = moles / concentration    volume = 0.1 mol / (0.002 mol/cm³)    volume = 50 cm³

Answer-ide.