The Ka : 1.671 x 10⁻⁷
<h3>Further explanation</h3>
Given
Reaction
HA (aq) + H2O (l) ←→ A- (aq) + H3O+ (aq).
0.3 M HA
pH = 3.65
Required
Ka
Solution
pH = - log [H3O+]
![\tt [H_3O^+]=10^{-3.65}=2.239\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20%5BH_3O%5E%2B%5D%3D10%5E%7B-3.65%7D%3D2.239%5Ctimes%2010%5E%7B-4%7D)
ICE method :
HA (aq) ←→ A- (aq) + H3O+ (aq).
0.3 0 0
2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴
0.3-2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴
![\tt Ka=\dfrac{[H_3O^+][A^-]}{[HA]}\\\\Ka=\dfrac{(2.239.10^{-4}){^2}}{0.3-2.239.10^{-4}}\\\\Ka=1.671\times 10^{-7}](https://tex.z-dn.net/?f=%5Ctt%20Ka%3D%5Cdfrac%7B%5BH_3O%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%5C%5C%5C%5CKa%3D%5Cdfrac%7B%282.239.10%5E%7B-4%7D%29%7B%5E2%7D%7D%7B0.3-2.239.10%5E%7B-4%7D%7D%5C%5C%5C%5CKa%3D1.671%5Ctimes%2010%5E%7B-7%7D)
Answer:
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Explanation:
Answer:
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Answer: 3M
Explanation: Molarity : It is defined as the number of moles of solute present in one liter of solution.
Formula used :
where,
n = moles of solute 
= 0.6 moles
= volume of solution in ml= 200 ml
Now put all the given values in the formula of molarity, we get
Therefore, the molarity of solution will be 3M.
The population decreases when the birth rate is less than the death rate.
