Chemistry: How many moles of HF (Ka=6.8×10−4) must be added to water to form 0.250 L of solution with a pH of 2.90?

How many moles of HF (Ka=6.8×10−4) must be added

marlesly87

18.01.2022

The number of moles of HF solution is 8.91 X10⁻⁴ mol

Explanation:

HF -------> H⁺ + F⁻

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

$K_a = \frac{\alpha^2.C_M}{(1-\alpha )}$

$[H^+] = \alpha .C_M$

$K_a = \frac{\alpha.[H^+]}{(1-\alpha )}$

$C_M = \frac{n(HF)}{V} = \frac{n(HF)}{0.25} = 4.n(HF)$

pH = -Log[H⁺]

$[H^+] = 10^{-pH} = 10^{-2.9} = 0.001259$

$K_a = \frac{\alpha.[H^+]}{(1-\alpha )} = \frac{\alpha(0.001259)}{(1-\alpha )} = 6.8 X 10^{-4}$

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, $[H^+] = \alpha .C_M$ , H⁺ = 0.001259 and Cm = 4n(HF)

$[H^+] = \alpha .C_M, 0.001259 = \alpha.4n(HF)$

$n(HF) = \frac{0.001259}{4\alpha} = \frac{0.001259}{4(0.3507)} = 0.000891 =8.91X10^{-4} mol$

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

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