Answer:
b. The weak base of an alkaline buffer will accept hydrogen protons when a strong acid is added to the solution
d.The conjugate acid of an alkaline buffer will donate hydrogen protons when a strong base is added to the solution.
Explanation:
A buffer is a solution that resist pH change, it shows minimal change upon addition of small amount of strong acid or strong base. An alkaline buffer will have a basic pH, above 7. It is made by mixing a weak base and its salt with a strong acid. An example of an alkaline buffer is carbonate-bicarbonate buffer that is prepared using varying amount of anhydrous sodium carbonate and volume of solution of sodium bicarbonate to get pH range between 9.2 to 10.7
Within the buffer,the salt is completely ionized while the weak base is partly ionized. on addition of acid, the released protons will be removed by the bicarbonate ion to form sodium carbonate; on addition of base, the hydroxide ion released by the base will be removed by the hydrogen ions to form water and the pH remains relatively the same
Answer:
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Explanation:
Hope this help!
Equation is as follow,
<span> 4 Na (s) + O</span>₂ <span>(g) → 2Na</span>₂<span>O (s)
According to equation,
91.92 g (4 moles) of Na produces = 123.92 g (2 moles) of Na</span>₂O
So,
17.4 g of Na will produce = X g of Na₂O
Solving for X,
X = (17.4 g × 123.92 g) ÷ 91.92 g
X = 23.45 g of Na₂O
The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
Read more on equilibrium constant;
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