Answer:
The number of moles of HF solution is 8.91 X10⁻⁴ mol
Explanation:
HF -------> H⁺ + F⁻
[tex]K_a = \frac{[H^+].[F^-]}{[HF]}[/tex]
[tex]K_a = \frac{\alpha^2.C_M}{(1-\alpha )}[/tex]
[tex][H^+] = \alpha .C_M[/tex]
[tex]K_a = \frac{\alpha.[H^+]}{(1-\alpha )}[/tex]
[tex]C_M = \frac{n(HF)}{V} = \frac{n(HF)}{0.25} = 4.n(HF)[/tex]
pH = -Log[H⁺]
[tex][H^+] = 10^{-pH} = 10^{-2.9} = 0.001259[/tex]
[tex]K_a = \frac{\alpha.[H^+]}{(1-\alpha )} = \frac{\alpha(0.001259)}{(1-\alpha )} = 6.8 X 10^{-4}[/tex]
α(0.001259) = 6.8 X 10⁻⁴ - (6.8 X 10⁻⁴)α
(0.001939)α = 0.00068
α = 0.00068/0.001939
α = 0.3507
Recall that, [tex][H^+] = \alpha .C_M[/tex], H⁺ = 0.001259 and Cm = 4n(HF)
[tex][H^+] = \alpha .C_M, 0.001259 = \alpha.4n(HF)[/tex]
[tex]n(HF) = \frac{0.001259}{4\alpha} = \frac{0.001259}{4(0.3507)} = 0.000891 =8.91X10^{-4} mol[/tex]
Therefore, the number of moles of HF solution that must be added to water to form 0.250 liters of solution with a pH of 2.9 is 8.91 X10⁻⁴ mol
