HF is a weakacid with a Ka of 7.1x10^-4. Now consider an aqueous solution made by...

HF is a weakacid with a Ka of 7.1x10^-4. Now consider an aqueous solution made by dissolving 0.210 grams of NaF in 250 mL of water. (NaF has a molar mass of 42.0 g/mol)
1. Write the equations for the processes that will take place when the NaF is dropped in water.
2. Calculate the pH in the solution described above.

Expert Solution

when NAF is dissolved in water, it undergoes ionization as shown below.

F-+ H2O----> OH- + HF

mole of NaF = mass/molar mass = 0.21/42=0.005, this is dissolved in 250ml (250/1000=0.25L) of water.

concentration of NaF= moles/Volume inL=0.005/0.25 =0.02M

for the reaction F-+ H2O ------->OH- + HF

Kb= [OH-] [HF]/[F-]=10^-14/Ka= 10^-14/ (7.1*10^-4)= 1.41*10^-11

preparing the ICE Table

component initial concentration (M) change Equilibrium concentration

F- 0.02 -x 0.02-x

HF 0 x x

OH- 0 x x

Kb= x²/(0.02-x)= 1.41*10^-11, when solved using excel, x= 5.31*10^-7, pOH= -log (5.31*10^-7)= 6.3

pH= 14-6.3= 8.7

queen_honey_blossom answered 2 years ago

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