Part 4: For region 5 in the graph above (the weak acid with more than an equival

Question

Part 4:

For region 5 in the graph above (the weak acid with more than an equivalent amount of strong base added), which of the following calculations would you do?

A. Ka = x2/[HA], pH = -log[H+]

B. pH = pKa + log([A-]/[HA])

C. pH = pKa

D. pOH = -log[OH-], pH = 14-pOH

E. Kb = x2/[A-], pOH = -log[OH-], pH = 14-pH

Part 5:

Suppose you have a 1.00 L solution with 0.448 moles of HF (Ka = 7.2*10-4). You have added 0.448 moles of NaOH to the solution. Using the diagram above, what numbered region of the titration have you reached?

Part 6:

Using the calculation method selected above, determine the pH of the solution in part 5 after the NaOH has been added. Note: consider the volume to remain unchanged.

Explanation / Answer

part 4

When more than one equivalent of strong base is added to the solution of weak acid, the solution contains a strong base and a salt.

Thus the pH of the solution sholud be calculated as

pH = 14-pOH and pOH = -log [OH-]

as the solution has more of Oh- ions.

part 5

When 1.00L of 0.448 M weak acid is added with 0.448moles of NaOH, the acid is exactly neutralised. Thus the region that corresponds to this point in the graph is point 4(equivalence point)

part 6

The pH of the solution at equivelnce point

HA + NaOH ----------> A- + H2O

0.448 0.448 0 0 inital

0 0 0.448 - at the equivalence

Thus [salt] = 0.448M

and pH is given by

pH = 1/2 [pKw +pKa + logC]

=1/2 [14 + (4-log7.2) + log 0.448]

= 8.3969

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects