Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
I would pick the first option in the third option
Answer:
The pressures will remain at the same value.
Explanation:
A catalyst is a substance that alter the rate of a chemical reaction. It either speeds up the or slows down the rate of a chemical reaction.
While a catalyst affects the rate, it is noteworthy that it has no effect on the equilibrium position of the chemical reaction. A catalyst works by creating an alternative pathway for the reaction to proceed. Most times, it decreases the activation energy needed to kickstart the chemical reaction.
Hence, we know that it has no effect on the equilibrium position. Factors affecting equilibrium position includes, temperature and concentration of reactants and products( pressure in terms of gases).
The reactants and the products here are gaseous, and as such pressure affects the equilibrium position. Now, we have established that the equilibrium position is unaffected. And as such the pressure affecting it does not change.
Thus, we have established that the pressure of the products and reactants are unaffected and as such they remain at their value unaffected.
