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?

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

Explanation:

HF -------> H⁺ + F⁻

pH = -Log[H⁺]

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⁺ = 0.001259 and Cm = 4n(HF)

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

i think its 5049.

i hope this !