It obtains a neutral pH (7). The combination of a strong acid and a strong base results in a neutral pH. The pH of HCl is about 1, and the pH of NaOH is 14.
Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
Answer:
8.61 mL of the HCl solution
Explanation:
The reaction that takes place is:
- 2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O
From the given mass of Mg(OH)₂, we can calculate <u>the moles of HCl that are neutralized</u>:
- 4x10² mg = 400 mg = 0.400g
- 0.400g Mg(OH)₂ ÷ 58.32g/1mol = 6.859*10⁻³ mol Mg(OH)₂
- 6.859*10⁻³ mol Mg(OH)₂ *
3.429x10⁻³ mol HCl
Finally, to calculate the volume of an HCl solution, we need both the moles and the concentration. We can <u>calculate the concentration using the pH value</u>:
= [H⁺]
- 0.0398 M = [H⁺] = [HCl] *Because HCl is a strong acid*
Thus, the volume is:
- 0.0398 M = 3.429x10⁻³mol HCl / Volume
- Volume = 8.616x10⁻³ L = 8.62 mL
