2. A) In the first beaker, we have a 0.600 M aqueous solution of HF. What is the

Question

2. A) In the first beaker, we have a 0.600 M aqueous solution of HF. What is the pH at equilibrium? Adapted from the University of Colorado at Boulder CHEM 1133 Recitation Manual, the Kenwood Academy Chemistry Department B) In a separate beaker, 0.500 g NaF was dissolved in water. What is the concentration of F? C) Now, combine the two beakers. Make a prediction about how the pH will change when F is added D) Calculate the pH of the new solution. Did the calculated pI match your prediction? Hint: Use the expression for Ka determined in the beginning of this problem.

Explanation / Answer

For part A)

We have the dissociation of HF in water:

HF <-> H+ + F-

We make our ICE table to determine equilibrium concentrations:

Using the definition of constant:

Ka = [H+][F-] / [HF]

6.7 x 10-4 = x2 / (0.6 - x)

Isolating for x:

x = [H+] = 0.01972 M

pH = -log [H+] = - log(0.01972) = 1.7

For part B)

Assuming a volume of 500 mL (if volume is specified just change the volume here):

First we turn grams into moles:

0.5 g * (1mol / 41.99 g) = 0.0119 moles

Molarity = 0.0119 moles / 0.5 L = 0.02381 M

As NaF has a 1:1 stoichiometry in its dissociation:

[F-] = 0.02381 M

Part C)

pH will increase as the adding of F- ions will decrease rate of dissociation.

Part D)

We go back to equilibrium stated in part A:

Using definition of constant:

6.7 x 10-4 = x*(0.02381+x) / (0.6 - x)

Isolating for x:

x = [H+] = 0.01125 M

pH = -log (0.01125) = 1.9488

Prediction confirmed

HF <-> H+ + F- I 0.6 M 0 0 C -x +x +x E 0.6 - x x x

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