Hydrogen pthalate ion, HC8H4O4-, is a weak acid and is produced when KHP (abbrev

Question

Hydrogen pthalate ion, HC8H4O4-, is a weak acid and is produced when KHP (abbrevation for KHC8H4O4) is dissolved in water. 1.02 g of KHP is dissolved in 100.0 mL of solution which is then titrated with NaOH solution of unknown concentration. The acid-base reaction that occurs is represented by the following net ionic equation:

HC8H4O4-(aq) + OH-(aq) ---> C8H4O4^2-(aq) + H20(l)

a. What is the initial molar concentration of HC8H4O4- before the titration?

b. During the titration 40,0 mL of NaOH solution were required to neutralize the acid. What is the molar concentration of NaOH?

c. At mid-point of titration the solution shows pH = 5.25. Calculate the Ka for hydrogen phthatlate ion, HC8H4O4-, and the kb of the phthalate ion, C8H4O4^-2.

d. Using the Kb and concentration of C8H4O4^2- at equivalence point, calculate the pH of the solution at the equivalence point. (Equivalence point is reached after the addition of 40.0 mL NaOH). Given: The following reaction occurs at equivalent point: C8H4O4^2-(aq) + H20(l) <---> HC8H4O4-(aq) + OH-(aq)

Explanation / Answer

a) initial moles of KHP = moles of HP- ion = mass of KHP/molar mass of KHP = 1.2/204.22 = 0.00588

HP- ion conc = moles/vol in L = 0.00588/0.1 = 0.0588 M

b) at neutralisation point moles of acid = moles of base

0.00588 = M x ( 40/1000)

M = 0.147 M

c) at mid point of titration pH = pka , hence pka = 5.25 , Ka = 10^-5.25 = 5.623 x 10^-6 for HP- ( Hydrogen phthalate ion)

we see that Ka ( HP-) x Kb ( P2- ion) = 10^-14

hence 5.623 x 10^-6 x Kb( pthalate ion) = 10^-14

Kb ( phthalate ion) = 1.78 x 10^-9

d) at equivalence point moles of C8H4O2- produced = moles of HP- ions initiaaly = 0.00588

at equi P2- + H2O <------> HP- + OH- (aq)

initially P2- moles = 0.00588 , HP-=OH-moles =x

thus at equi P2- moles = 0.0588-x , HP- moles = x =OH- moles

Kb =[OH-][HP-]/[P2-]

where vol = 100 ml + 40 ml = 140 ml = 0.14 L

Kb = 1.78 x 10^-9 = ( x/0.14)(x/0.14) /( 0.00588-x)/0.14

x = 1.2 x 10^-6 moles =OH- moles

[OH-] = 1.2 x 10^-6 /0.14 = 8.646 x 10^-6

pOh = -log ( 8.646 x 10^-6) = 5.06

pH = 14-5.06 = 8.94

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

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