Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka= ![\frac{ [H^{+}] [A^{-} ]}{ [HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BH%5E%7B%2B%7D%5D%20%20%5BA%5E%7B-%7D%20%5D%7D%7B%20%5BHA%5D%7D)
Ka= 
Ka= 
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka= 
= 
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶
As a base is added to an acidic solution, the H+ ions in solution that make it acidic are slowly neutralized into water (via OH-, the base). As these ions are converted into water the concentration of them decreases, so the pH decreases, as they are directly related.
Hope this helps!
Answer:
The answer to your question is: letter D
Explanation:
In a combustion reaction, the reactants are always a molecule with Carbon that reacts with oxygen and the products are carbon dioxide and water.
According to the explanation, the only possible solution is:
a) C₆H₁₂O₂(l) ⇒ 6 C(s) + 6 H₂(g) + O₂(g)
b) Mg(s) + C₆H₁₂O₂(l) ⇒ MgC₆H₁₂O₂(aq)
c) 6 C(s) + 6 H₂(g) + O₂(g) ⇒ C₆H₁₂O₂(l)
d) C₆H₁₂O₂(l) + 8 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(g)
e) None of the above represent the combustion of C₆H₁₂O₂.
Answer:
The solution is basic.
Explanation:
We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):
- If no. of millimoles of acid > that of base; the solution is acidic.
- If no. of millimoles of acid = that of base; the solution is neutral.
- If no. of millimoles of acid < that of base; the solution is basic.
- We need to calculate the no. of millimoles of acid and base:
no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.
no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.
<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>
<em>So, the solution is: basic.</em>
