To
determine the percent ionization of the acid given, we make use of the acid
equilibrium constant (Ka) given. It is the ration of the equilibrium
concentrations of the dissociated ions and the acid. The dissociation reaction
of the HF acid would be as follows:
HF = H+ + F-
The acid equilibrum constant would be expressed as follows:
Ka = [H+][F-] / [HF] = 3.5 x 10-4
To determine the equilibrium concentrations we use the ICE table,
HF
H+ F-
I 0.337 0
0
C -x +x
+x
---------------------------------------------
E 0.337-x x
x
3.5 x 10-4 = [H+][F-] / [HF]
3.5 x 10-4 = [x][x] / [0.337-x]
Solving for x,
x = 0.01069 = [H+] = [F-]
percent ionization = 0.01069 / 0.337 x 100 = 3.17%
The percent ionization of a 0.337 M HF solution : 3.2%
According to Arrhenius, acids are substances which, when dissolved in water, release Hions.
An HₓY acid in water will ionize:
HₓY (aq) --------> xH⁺ (aq) + Yˣ- (aq)
Example:
HCl -------> H⁺ + Cl⁻
The amount of Hions produced by 1 acid molecule is called valence acid, whereas acidic residuals are formed after the release of Hions.
Usually, the name acid begins with the word acid followed by the name of the remaining acidic ion
The ion concentration of a weak acid is determined by the value of the acid ionization constant (Ka).
The greater the value of Ka, the greater the dissociated acid produces its Hion and the greater its acidity
HF is a weak acid
Weak acid ionization reaction occurs partially (not ionizing perfectly as in strong acids)
The ionization reaction of a weak acid is an equilibrium reaction
HA (aq) ---> H + (aq) + A- (aq)
The equilibrium constant for acid ionization is called the acid ionization constant, which is symbolized by Ka
The values for the weak acid reactions above:
The greater the Ka, the stronger the acid, which means the reaction to the right is also greater
The degree of ionization (symbol α) is the ratio of the amount of ionizing substance to the amount of substance dissolved
α = amount of substance ionizing : amount of substance dissolved
HF decomposition reaction
HF ---> H⁺+ F⁻ if there is 0.337 M HF, then
[tex]\rm Ka=\dfrac{[H^+][F^-]}{[HF]}[/tex]
3.5 x 10⁻⁴. = x. x : 0.337-x (0.337-x is considered proportional to 0.337 because x is very small)
1,179.10⁻⁴ = x²
x = 0.01086
α = 0.01086: 0.337
α = 0.0322
α = 3.22%
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