Question

Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions. (Ka(HF)=6.8×10^−4

A. 0.13 MM HF

B. 0.13 MM NaF

C. a mixture that is 0.13 MM in HFHF and 0.13 MM in NaF

Answer 1

- 0.13 mg Hf wat Httf Ka6.8x1024 HG 2 -0.13M 10.13-a n a 6.8x10-4 = x 0.13-x 0.13-1 0 .13. 21 = 0.8 x 0.13 Xlony z = 0.884 X109 olis = 0,940X102 pH = -log 10.990 81002) = = 2*0.026 2.026 6 0.13M Naf Naf ff- ft H₂O 0.13 -2 Hft OH Q : 10-19 ------03- Scanned with CamScanner

a² = 0.13*0.197x10-10 20 = 0.0190 X10-10 me = 0.138 X10-5 POH = 5 + 0.86 5-86 pH = 14-5.86 = 8:14 0 0.13M Mf and 0.13m Naf Buffer solution pH = pka + log [Naf pH = 3.167 + log (1) pH = 3.167 Scanned with CamScanner