Answer: The pH of 0.10 M 
is 4.49.
Explanation:
Given: Initial concentration of 
= 0.10 M
Let us assume that amount of dissociates is x. So, ICE table for dissociation of is as follows.
![Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}](https://tex.z-dn.net/?f=Cu%28H_%7B2%7DO%29%5E%7B2%2B%7D_%7B6%7D%20%5Crightleftharpoons%20%5BCu%28H_%7B2%7DO%29_%7B5%7D%28OH%29%5D%5E%7B%2B%7D%20%2B%20H_%7B3%7DO%5E%7B%2B%7D)
Initial: 0.10 M 0 0
Change: -x +x +x
Equilibrium: (0.10 - x) M x x
As the value of 
is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.
And, (0.10 - x) will be approximately equal to 0.10 M.
The expression for value is as follows.
![K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BCu%28H_%7B2%7DO%29%5E%7B2%2B%7D_%7B6%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BCu%28H_%7B2%7DO%29%5E%7B2%2B%7D_%7B6%7D%5D%7D%5C%5C1.0%20%5Ctimes%2010%5E%7B-8%7D%20%3D%20%5Cfrac%7Bx%20%5Ctimes%20x%7D%7B0.10%7D%5C%5Cx%20%3D%203.2%20%5Ctimes%2010%5E%7B-5%7D)
Hence, ![[H_{3}O^{+}] = 3.2 \times 10^{-5}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%203.2%20%5Ctimes%2010%5E%7B-5%7D)
Formula to calculate pH is as follows.
![pH = -log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D)
Substitute the values into above formula as follows.
![pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5C%3D%20-%20log%20%283.2%20%5Ctimes%2010%5E%7B-5%7D%29%5C%5C%3D%204.49)
Thus, we can conclude that the pH of 0.10 M is 4.49.
Answer:
-1.71 J/K
Explanation:
To solve this problem we use the formula
ΔS = n*ΔH/T
Where n is mol, ΔH is enthalpy and T is temperature.
ΔH and T are already given by the problem, so now we calculate n:
Molar Mass C₂H₅OH = 46 g/mol
2.71 g C₂H₅OH ÷ 46g/mol = 0.0589 mol
Now we calculate ΔS:
ΔS = 0.0589 mol * −4600 J/mol / 158.7 K
ΔS = -1.71 J/K
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
