Question

What is the percent dissociation of a 0.400 M HF solution its pH = 2.08?

What is the K_a value HF at this temperature?

Answer 1

Answer 2

 

1. Calculate the Percent Dissociation of HF

Given:

• Initial concentration of HF, $[HF{]}_0=0.400\text{M}$

• pH of the solution = 2.08

Step 1: Find $[H^{+}]$ from pH

$$[H^{+}]=10^{-\text{pH}}=10^{-2.08}\approx 8.32\times 10^{-3}\text{M}$$

Step 2: Determine the Dissociated Concentration of HF
Since HF dissociates as:

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

The concentration of $[H^{+}]$ at equilibrium is equal to the dissociated concentration of HF:

$$[\text{HF}{]}_{\text{dissociated}}=[H^{+}]=8.32\times 10^{-3}\text{M}$$

Step 3: Calculate Percent Dissociation

$$\text{Percent Dissociation}=\left(\frac{[\text{HF}{]}_{\text{dissociated}}}{[\text{HF}{]}_0}\right)\times 100=\left(\frac{8.32\times 10^{-3}}{0.400}\right)\times 100\approx 2.08\mathrm{\%}$$

Answer:
The percent dissociation of the 0.400 M HF solution is 2.08%.

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2. Calculate the $K_a$ of HF

Given:

• $[H^{+}]=8.32\times 10^{-3}\text{M}$

• $[F^{-}]=[H^{+}]=8.32\times 10^{-3}\text{M}$ (from dissociation)

• $[HF{]}_{\text{equilibrium}}=[HF{]}_0-[H^{+}]=0.400-8.32\times 10^{-3}\approx 0.3917\text{M}$

Step 1: Write the $K_a$ Expression

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

Step 2: Substitute Values

$$K_a=\frac{(8.32\times 10^{-3})(8.32\times 10^{-3})}{0.3917}\approx \frac{6.92\times 10^{-5}}{0.3917}\approx 1.77\times 10^{-4}$$

Answer:
The $K_a$ value of HF at this temperature is $1.77\times 10^{-4}$