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
2LiOH + H₂SO₄ = Li₂SO₄ + 2H₂O
v(H₂SO₄)=0.025 L
v(LiOH)=0.0836 L
c(LiOH)=0.12 mol/L
n(LiOH)=v(LiOH)c(LiOH)
n(H₂SO₄)=n(LiOH)/2=v(LiOH)c(LiOH)/2
c(H₂SO₄)=n(H₂SO₄)/v(H₂SO₄)=v(LiOH)c(LiOH)/(2v(H₂SO₄))
c(H₂SO₄)=0.0836*0.12/(2*0.025)=0.20064 mol/L
≈0.20 M
Answer:
It's neutral
Explanation:
Pure water is neither acidic or basic; so it is neutral
Answer:
a. True
Explanation:
Methyl red is an indicator widely used in quality control of oxides as Zinc oxide in the titration with sulfuric acid.
As is used in titrations of acid-base reactions the indicator change in colour. Is red when the pH < 4.4 (Acidic Solutions) and is yellow when pH > 6.2 (Neutral-Basic solutions).
A change in colour means the structure of the indicator is changing with pH. Thus, the answer is:
<h3>a. True
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