Answer:
The pH of the 0.100 M solution of hydobromous acid HBrO is 4.843
Explanation:
Here we have the reaction given as follows;
HBrO ⇄ H⁺ + BrO⁻¹
Therefore;
HBrO ⇄ H⁺ + BrO⁻¹
Initial concentration of HBrO = 0.100 M decomposes partly to form x moles of each of H⁺ and BrO⁻¹. That is
HBrO ⇄ H⁺ + BrO⁻¹
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1 - x x x
Hence;
![Ka = \frac{[H^+][BrO^{-1}]}{[HBrO]} =\frac{x \times x }{0.1 - x} = \frac{x^2}{0.1 -x} = 2.06 \times 10^{-9}](https://tex.z-dn.net/?f=Ka%20%3D%20%20%5Cfrac%7B%5BH%5E%2B%5D%5BBrO%5E%7B-1%7D%5D%7D%7B%5BHBrO%5D%7D%20%3D%5Cfrac%7Bx%20%5Ctimes%20x%20%7D%7B0.1%20-%20x%7D%20%3D%20%5Cfrac%7Bx%5E2%7D%7B0.1%20-x%7D%20%3D%202.06%20%5Ctimes%2010%5E%7B-9%7D)
(0.1 - x) × 2.06×10⁻⁹ = x²
x² + 2.06×10⁻⁹·x - 0.1 = 0
Factorizing gives;
(x + 1.4354×10⁻⁵)(x - 1.4352×10⁻⁵) = 0
Therefore, x = 1.4352×10⁻⁵ M or -1.4354×10⁻⁵ M
We take the positive value as x is the concentration of the ions in the solution;
From the above [H⁺] = 1.4352×10⁻⁵ M
pH = -log[H⁺] = -log(1.4352×10⁻⁵) = 4.843
The pH of the 0.100 M solution of hydobromous acid HBrO at 298 K where the Ka = 2.06×10⁻⁹ = 4.843.
I believe the correct answer is A
Answer:
1) a) 2 C2H2 + 3O2 --> 2CO2 + 2H2O
b) Combustion Reaction
c) Bonds will be broken on the reactant side which will require energy.
d) It agrees with the law of conservation of mass because there are the same amount of molecules of each element on both sides of the equation, meaning mass was not created nor destroyed.
2) a) 2Na + Cl2 --> 2NaCl
Synthesis
b) 2Fe + O2 --> 2FeO
Synthesis
Hope this helps!
Explanation:
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