3. Suppose that part (II) of the experiment were carried out with aqueous NaOH instead of Ba(OH)2 to form soluble Na2SO4 along with H2O as products.
(a) Write the overall balanced equation.
(b) Write the net ionic equation.
(c) How would the graph of conductivity versus volume of aqueous NaOH added differ from the graph you constructed using Ba(OH)2?
To answer this question, we need to consider the reactants and products involved in the reaction between NaOH and H2SO4. Let's break it down step by step:
(a) Writing the overall balanced equation:
The reaction between NaOH and H2SO4 produces Na2SO4 and H2O. The balanced equation can be written as follows:
2 NaOH + H2SO4 -> Na2SO4 + 2 H2O
(b) Writing the net ionic equation:
To write the net ionic equation, we need to consider the dissociation of the involved compounds in water. NaOH and H2SO4 will dissociate into their respective ions in solution. However, Na2SO4 is a soluble salt and will remain dissociated in water. The net ionic equation is as follows:
2 OH- + H+ -> 2 H2O
(c) Difference in the graph of conductivity versus volume of NaOH added:
In the previous experiment with Ba(OH)2, the formation of a precipitate (BaSO4) caused a significant increase in conductivity, indicating the presence of ions in solution. However, in this new experiment with NaOH, Na2SO4 formed is soluble in water, meaning no precipitate will be formed. As a result, there will be no significant change in conductivity as NaOH is added. The graph for the NaOH experiment will show a much less pronounced increase in conductivity compared to the Ba(OH)2 experiment.
Remember, to obtain accurate and complete answers, it's important to have a good understanding of chemistry concepts and the ability to apply them to specific situations.