Question

What is the pH of a solution made by dissolving 5.06 grams of calcium fluoride in enough water to make 7.0×10² mL of solution? The K_a for HF is 6.8x10^–4.

Answer 1

Giren, - Mass of Cafe= 5.069 Molar mass of Cafe = 78.07 g/mol ... No. of Moles of lafz = 0.065 moles In 7.0x10²mL of solution, no. of moles = 0.065 moles ... In 1000 mL of solution, no. of moles = 0.065 x1000 #x202 = 0.093 :Concentration of car, - 0.093(M) . Cat will dissociate as - CaF2 ~ Ca 2+ + af Initial conc. 0.093 0 0 Arequilibriem 0.093 2 x 0.093 = 0.186M cs Scanned with with oncentration er if = 0.186 M Cam Seanne concentua

To find the pH of the solution, it is governed by the equation - f- Hot HF + oH k = [Hf][own] [F-] Two other equilibria also exist, H2O = H,+ oH (1),kw = [H"][or] HF + H₂O ft H₂O + (2), ka?[E] [₂0] - [hf Now, kw - CHF] [ou] ka [f- - k given ka= 6.8x10-4 ko = 10-14 at 25°c ik: 1.471x10*" Now, [OH-] [ HF] = k.[f.] Assuming [ou ] = [H.] (from the stoichiometry of the reaction), then : [OH-] = Sk. [F-] - 1.471x10 -"x 0.186 cs Scanned with $1.654x10-6 CamScanner

- p0Hz- log [omi] -- log (1.654X106) = 5.78 Now, PH+ pOH = 14 ii DH = 14-5.78 = 8.218 ...ph of the solution is 8.218. / Scanned with CamScanner