Question

Calculate the percent ionization of formic acid solutions having the following concentrations.a) 1.20Mb)0.450Mc)0.130Md)5.10 x 10^-2

Answer 1

Answer:

a) 1.224 %.

b) 2.0 %.

c) 3.7 %.

d) 5.94 %.

Explanation:

• To solve this problem, we have the relation:

Ka = (α²C)/(1 - α),

where, Ka is the Dissociation Constant of the acid; C is the Concentration of the acidic solution;  and α is the degree of dissociation of the acid.

• Since, α is a very small value, (1 - α) ≅ 1.

∴ Ka = (α²C).

∴ α = √(Ka/C).

a) 1.20 M:

∵ α = √(Ka.C)

Ka of formic acid = 1.8 x 10⁻⁴, C = 1.2 M.

∴ α = √(Ka/C) = √(1.8 x 10⁻⁴)/(1.2) = 0.01224.

∴ The percent of ionization = α x 100 = 1.224 %.

b) 0.45 M:

∵ α = √(Ka.C)

Ka of formic acid = 1.8 x 10⁻⁴, C = 0.45 M.

∴ α = √(Ka/C) = √(1.8 x 10⁻⁴)/(0.45) = 0.02.

∴ The percent of ionization = α x 100 = 2.0 %.

c) 0.13 M:

∵ α = √(Ka.C)

Ka of formic acid = 1.8 x 10⁻⁴, C = 0.13 M.

∴ α = √(Ka/C) = √(1.8 x 10⁻⁴)/(0.13) = 0.037.

∴ The percent of ionization = α x 100 = 3.7 %.

d) 5.10 x 10⁻² M:

∵ α = √(Ka.C)

Ka of formic acid = 1.8 x 10⁻⁴, C = 5.10 x 10⁻² M.

∴ α = √(Ka/C) = √(1.8 x 10⁻⁴)/(5.10 x 10⁻²) = 0.0594.

∴ The percent of ionization = α x 100 = 5.94 %.

Answer 2

a). 1.22%

b). 2%

c). 3.72%

d). 5.94%

Further explanation

Given:

Ka for formic acid, HCOOH, is $1.80 \times 10^{-4}$

Question:

Calculate the percent ionization (α) of formic acid solutions having the following concentrations (M).

The Process:

Remember that,

$\boxed{ \ [H^+] = \alpha M \ }$ and $\boxed{ \ [H^+] = \sqrt{K_a M} \ }$

The two equations are rearranged to connect α, Ka, and M directly.

$\boxed{ \ \alpha M = \sqrt{K_a M} \ }$

$\boxed{ \ (\alpha M)^2 = (\sqrt{K_a M})^2 \ }$

$\boxed{ \ \alpha^2 M^2 = K_a M \ }$

$\boxed{ \ \alpha^2 = \frac{K_a M}{M^2} \ }$

$\boxed{\boxed{ \ \alpha = \sqrt{\frac{K_a}{M}} \ }}$

Let us calculate the percent ionization from each concentration below.

a). 1.20 M

$\boxed{ \ \alpha = \sqrt{\frac{1.80 \times 10^{-4}}{1.20}} \ }$

$\alpha \approx 0.0122$

Thus, the percent of ionization for 1.20 M HCOOH solution is 1.22%.

_ _ _ _ _ _ _

b). 0.450 M

$\boxed{ \ \alpha = \sqrt{\frac{1.80 \times 10^{-4}}{0.450}} \ }$

$\alpha \approx 0.02$

Thus, the percent of ionization for 0.450 M HCOOH solution is 2%.

_ _ _ _ _ _ _

c). 0.130 M

$\boxed{ \ \alpha = \sqrt{\frac{1.80 \times 10^{-4}}{0.130}} \ }$

$\alpha \approx 0.0372$

Thus, the percent of ionization for 0.130 M HCOOH solution is 3.72%

_ _ _ _ _ _ _

d). 5.10 x 10⁻² M

$\boxed{ \ \alpha = \sqrt{\frac{1.80 \times 10^{-4}}{5.10 \times 10^{-2}}} \ }$

$\alpha \approx 0.0594$

Thus, the percent of ionization for 5.10 x 10⁻² M HCOOH solution is 5.94%.

_ _ _ _ _ _ _ _ _ _

Notes

Just a reminder, Ka and α are characteristic of weak acid solutions.

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