a. A 250.0 mL buffer solution contains 0.157 M RbF and 0.165 M HF. For HF, Ka= 3

Question

a. A 250.0 mL buffer solution contains 0.157 M RbF and 0.165 M HF. For HF, Ka= 3.5x10^-4

Label each as a strong or weak acid ;strong or weak base; acidic, basic or neutral salt: HF= _________RbF= ________ HBr= ___________

b. Write the hydrolysis reaction for HF (aq).

c. Can the Henderson Hasselbach equation be used to calculate the pH? Why or why not?

d. Calculate the pH of the buffer solution from part a. Show work!

e. Write the net-ionic neutralization reaction that occurs when HBr (aq) is added to the buffer solution in part a.

f. What is the pH after 75.0 mL of 0.195 M HBr (aq) is added to the 250.0 mL buffer solution in part a? Show your calculations for the initial moles of HF, RbF and HBr and set up a change table for the reaction written in part e Also show your pH calculation.

Explanation / Answer

a)

RbF --> Rb+ + F-

F- hydrolyses with water to form

F- + H2O <-> HF + OH-

then, it is a weak base

HBr --> H+ +Br- in 100% since strong acid

b)

For HF

HF(aq) + H2O(l) <-> H3O+(aq) + F-(aq)

c)

it can only be used when there is conjguate aicd + base

i.e. F- and HF

d)

the pH can be clauclated via

pH = pKa + log(F-/HF)

pKa = -log(Ka) = -log(3.5*10^-4) = 3.45

pH = 3.45 + log(0.157/0.165)

pH = 3.4284

e)

net ionic neutralikzation

HBr = H+ + Br-

F- + H+ <--> HF

then

HBr + F- --> HF + Br-

is the net ionic

f)

mmol of HF= 0.165*250 = 41.25

mmol of F- = MV = 0.157*250 =39.25

after H+ addition

mmo of H =MV = 0.195*75 = 14.625

mmol of HF formed = 14.625 + 41.25 =55.875

mmol of F- = 39.25-14.625 = 24.625

p H= 3.4 + log(24.625/55.875)

pH = 3.0441

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.