Find the percent ionization of a 0.337 m hf solution. the ka for hf is 3.5 x 10-4. 1.1 % 1.2 x 10-2 % 3.2 % 3.5 x 10-2 % 4.7 %

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

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%

calculate kc

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i think it's letter a.