[LunarFicus]

Hi, I need help with the following problem:

In a 500mL beaker, 65.0g NH4F is dissolved in 750.mL distilled water. Determine the pH of
the solution. Kb (NH3) = 1.8x10^-5 Ka (HF) = 7.4x10^-4

I tried finding the molarity of the concentration by finding the Moles (Grams of NH4F/Grams per mole of NH4F) of NH4F being dissolved and dividing that by the liters of solution. Using the molarity I found the hydronium ion concentration of NH4F and plugged it into -log[H+] to find the pH. However, I don't know if this method is correct because I didn't have to use the Ka and Kb values provided in the question. What is the right way to solve this?

[hypervalent_iodine]

What is the right way to solve this?

By using the Ka / Kb values of ammonia / the ammonium ion. NH₄F is not a strong acid, so [NH₄⁺] ≠ [H⁺]. Do you know how to construct ICE tables?

[LunarFicus]

Yeah, you put the initial M values, the change in those values, and then the end result right?

[hypervalent_iodine]

Yeah, you put the initial M values, the change in those values, and then the end result right?

More or less.

[AWK]

In a 500mL beaker, 65.0g NH4F is dissolved in 750.mL distilled water

?
http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified