Question

This question has two parts: What is the pH of a 0.1 M solution of HF? b) What is the percent dissociation of a 0.1 M HF solution? PLEASE SHOW WORK TYPED. It's hard to understand some handwritings so this is the easiest way I can ensure I can actually read the answer and understand it.

Answer 1

Solution:

HF is a weak acid, hence it dissociated as,

HF = H+ + F-

0.10---0-----0 (initial concentration)

(0.10- X)---X---X (at equilibrium)

Dissociation constant, Ka = X x X / (0.10 - X)

Ka for HF = 3.5 x 10^-4

Therefore, 3.5 x 10^-4 = X2 / 0.10-X

X2 + 3.5 x.10^-4 X - 3.5 x 10^-5 = 0

On solving the quadratic equation,

X = 5.74 x 10^-3 M

pH = -log X

pH = -log ( 5.74 x 10^-3)

pH = 2.24

Part B)

Percentage dissociation =( X/ 0.10 M ) x 100

= 5.74 x10^-3 x 100/ 0.10 = 5.74 %

Answer 2

Given:

• Concentration of HF ($C$) = 0.1 M

• Acid dissociation constant ($K_a$) of HF = 6.6 × 10⁻⁴

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Part a).  pH of the Solution :-

1. Dissociation Equation:

   $$\text{HF}\rightleftharpoons {\text{H}}^{+}+{\text{F}}^{-}$$

2. Equilibrium Expression:

   $$K_a=\frac{[{\text{H}}^{+}][{\text{F}}^{-}]}{[\text{HF}]}$$

3. Let $x$ = [H⁺] at equilibrium:

• Initial:

  $$[\text{HF}]=0.1\text{M},[{\text{H}}^{+}]=0,[{\text{F}}^{-}]=0$$

• Change:

  $$-x,+x,+x$$

• Equilibrium:

  $$[\text{HF}]=0.1-x,[{\text{H}}^{+}]=x,[{\text{F}}^{-}]=x$$

4. Substitute into $K_a$:

   $$6.6\times 10^{-4}=\frac{x\cdot x}{0.1-x}\approx \frac{x^2}{0.1}$$

   (Approximation: $x\ll 0.1$, so $0.1-x\approx 0.1$)

5. Solve for $x$:

   $$x^2=6.6\times 10^{-5}⟹x=\sqrt{6.6\times 10^{-5}}\approx 0.0081\text{M}$$

   (This is the [H⁺] at equilibrium.)

6. Calculate pH:

   $$\text{pH}=-\log [{\text{H}}^{+}]=-\log (0.0081)\approx 2.09$$

 Answer:
The pH of the 0.1 M HF solution is 2.09.

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Part b). Percent Dissociation :-

1. Formula:

   $$\mathrm{\%}\text{Dissociation}=\left(\frac{[{\text{H}}^{+}]}{[\text{HF}{]}_{\text{initial}}}\right)\times 100$$

2. Substitute [H⁺] = 0.0081 M and [HF]₀ = 0.1 M:

   $$\mathrm{\%}\text{Dissociation}=\left(\frac{0.0081}{0.1}\right)\times 100=8.1\mathrm{\%}$$

Final Answer:
The percent dissociation of the 0.1 M HF solution is 8.1%.

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Answer 3

Given Data:

• Concentration of HF, $C = 0.1 \, \text{M}$

• $K_a$ for HF = $6.6 \times 10^{-4}$

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(a) To calculate the pH:

HF is a weak acid, so it dissociates partially:

$$\text{HF} \rightleftharpoons \text{H}^+ + \text{F}^-$$

Set up an ICE table:

Species	Initial (M)	Change (M)	Equilibrium (M)
HF	0.1	-x	0.1 - x
H⁺	0	+x	x
F⁻	0	+x	x

Apply the expression for the acid dissociation constant:

$$K_a = \frac{[H^+][F^-]}{[HF]}$$$$6.6 \times 10^{-4} = \frac{x^2}{0.1 - x}$$

Since $K_a$ is relatively small, we can approximate by assuming $0.1 - x \approx 0.1$, simplifying the equation:

$$6.6 \times 10^{-4} = \frac{x^2}{0.1}$$$$x^2 = (6.6 \times 10^{-4}) \times 0.1 = 6.6 \times 10^{-5}$$$$x = \sqrt{6.6 \times 10^{-5}} \approx 8.13 \times 10^{-3} \, \text{M}$$$$\text{pH} = -\log(8.13 \times 10^{-3}) \approx 2.09$$

✅ pH ≈ 2.09

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(b) To calculate the percent dissociation:

$$\text{Percent dissociation} = \frac{[H^+]}{[HF]_{\text{initial}}} \times 100$$$$\text{Percent dissociation} = \frac{8.13 \times 10^{-3}}{0.1} \times 100 \approx 8.13\%$$

✅ Percent dissociation ≈ 8.13%