Need Help With the Chemistry Aqueous Equilibria problem?

Question

1. Determine the pH of a solution prepared by adding 17.2mL of 0.157MKOH to 20.0mL of 0.132MHNO3 solution. 

2. Consider the titration of 15.0 mL of 0.0738M hydroazoic acid (HN3, Ka= 1.9x10^-5) 

(a). Determine the pH after 4.00mL of 0.108M NaOH has been added. 

(b). Determine the pH after a total of 12.0 mL of 0.108M NaOH has been added. 

3. Consider the following equilibrium: BaF2(s)-----> Ba^2+(aq) + 2F^–(aq) Ksp= 1.0 x 10-6 

Will BaF2 precipitate from a solution that is 0.015 M Ba^2+and 7.5x10-3M F^–? (Calculate Q and compare its value to Ksp.)

Answer 1

1.) KOH + HNO₃ ==> H₂O + KNO₃ ... balanced equation

moles KOH = 0.0172 L x 0.157 moles/L = 0.00270 moles

moles HNO₃ = 0.0200 L x 0.132 mol/L = 0.00264 moles

KOH is in excess by 0.00270 - 0.00264 = 0.00006 moles = 6x10^-5 moles

Total volume = 17.2 ml + 20.0 ml = 37.2 ml = 0.0372 L

Final [KOH] = 6x10^-5 moles/0.0372 L = 1.6x10^-3 M

pOH = -log [OH-] = -log 1.6x10^-3 = 2.80

pH = 14 - pOH = 14 - 2.80

pH = 11.2

2.) Hydrazoic acid is a weak acid and thus creates a buffer when titrated with NaOH.

HN₃ + NaOH ==> H₂O + NaN₃

a) before addition of NaOH; moles HN3 = 0.015 L x 0.0738 mol/L = 0.00111 moles HN3

moles NaOH added = 0.004 L x 0.108 mol/L = 0.000432 moles NaOH = 0.000432 moles NaN3 formed

moles HN3 left over = 0.00111 - 0.000432 = 0.000678 moles NH3

Final volume = 15 ml + 4 ml = 19 ml = 0.019 L

Final [NH3] = 0.000678 moles/0.019 L = 0.0357 M

Final NaN3] = 0.000432 moles/0.019 L = 0.0227 M

pH = pKa + log [NaN3]/[HN3] and pKa = -log Ka = -log 1.9x10^-5 = 4.721

pH = 4.721 + log (0.0227/0.0357) = 4.721 + (-0.197)

pH = 4.52

b) Do the same as above procedure correcting for moles of each and total volume

3.) BaF₂ ==> Ba²⁺ + 2F^- Ksp = 1.0x10^-6

Q = [Ba²⁺][F-]² = (0.015)(7.5x10^-3)² = 8.44x10^-7

Q is < Ksp so there will be no precipitate formed