Chemistry 溫習題目 9
1. The strength of acids can be determined by adding lumps of magnesium carbonate to the acids.
a) Write an ionic equation for the reaction between acid and magnesium carbonate.
b) Outline an experiment to determine the strengths of 0.1M sulphuric acid and 0.1M sulphurous acid using lumps of magnesium carbonate.
c) State and explain the expected experimental results in b).
2. When 2.0M NaOH(aq) is added to 20.0cm^3 of 2.0M HCl(aq), the changes in temperature of the reaction mixture are shown in the following table.
Volume of NaOH(aq) added (cm^3) / 0.0 / 5.0 / 10.0 / 15.0 / 20.0 / 25.0 / 30.0
Temperature of reaction mixture (C) / 20.0 / 26.0 / 31.0 / 35.0 / 38.0 / 32.0 / 20.0
a) Explain the change in temperature of the reaction mixture throughout the experiment.
b) Write an ionic equation for the reaction involved.
c) Determine the maximum temperature rise of the reaction mixture.
d) Predict, with explanation, the maximum temperature rise of the reaction mixture if
(i) the concentrations of the acid and alkali are doubled while the volumes of the solutions used remain the same, and
(ii) the volumes of the acid and alkali are doubled while the concentrations of the solutions used remain the same.
呢條我 plot 咗 graph，但係跟著唔識做...
- polarbearhmhLv 46 年前最愛解答
a) 2H⁺(aq) + MgCO₃(s) → Mg²⁺(aq) + H₂O(l) + CO₂(g)
(1) Add equal masses of magnesium carbonate separately to excess and equal volumes of the sulphuric acid and sulphurous acid.
(2) The acid that gives out colourless gas bubbles more quickly is the stronger acid.
(1) 0.1 M sulphuric acid gives out colourless gas bubbles more quickly than 0.1 M sulphurous acid.
(2) Sulphuric acid is a stronger acid than sulphurous acid.
(3) For the same molar concentration of acids, sulphuric acid contains more hydrogen ions for the reaction with magnesium carbonate.
(4) It reacts with magnesium carbonate more quickly to give carbon dioxide.
Neutralization is an exothermic reaction. When NaOH(aq) reacts with HCl(aq), temperature of the mixture rises. When the reaction just completes, the temperature reaches its maximum. Then the temperature starts to fall as addition of NaOH(aq) will not give out heat but cool the mixture.
b) H⁺(aq) + OH⁻(aq) → H₂O(l)
c) (By plotting the graph and joining the points before and after the reaction completes into two straight lines. The intersection point of the two lines is the point where maximum temp. is gained/the reaction completed. Below I will do it in a mathematical way.)
When NaOH(aq) reacts with HCl(aq), rate of change of temp. ≈ (38-20)/20 = 0.9°C cm⁻³; after the reaction completed, rate of change of temp. ≈ (20-32)/(30-25) = -2.4°C. Let the maximum temp. and volume of NaOH reacted be t°C and x cm³ respectively.
Clearly the reaction completed when volume of NaOH(aq) added was between 20.0 and 25.0cm³, then
(t-38)/(x-20) = 0.9
(32-t)/(25-x) = -2.4.
By solving the equation, t ≈ 39.6. Maximum temp. rise ≈ 39.6 - 20.0 = 19.6°C
d) Please refer to the comments.
2015-06-14 18:19:45 補充：
(i) The amount of heat released is doubled while the heat is used to heat the same volume of the solution. Therefore, the maximum temperature rise will be doubled.
2015-06-14 18:19:51 補充：
(ii) The amount of heat released is doubled since the amounts of hydrogen and hydroxide ions are doubled. While the volume of the solution that the heat is used to heat is also doubled. Hence the maximum temperature rise will remain unchanged.