A 50.0-mL solution of 0.052 M potassium hydrogen phthalate (KHC_gH_O_4; abbrevia

Question

A 50.0-mL solution of 0.052 M potassium hydrogen phthalate (KHC_gH_O_4; abbreviated as "KHP") is titrated with 0.10 M NaOH. (a) How many moles of KHP were titrated? (b) How many milliliters (mL) of 0.10 M NaOH solution are needed to reach the equivalent point according to the following net ionic equation? What is the molar concentration of the conjugate base C_8H_4O_4^2- (the phthalate ion) when the titration has reached equivalent point? a If the solution at equivalent point has pH = 8.96, calculate the K_b value for the conjugate base C_8H_4O_4^2- ? (e) What is the K_a value for hydrogen phthalate ion, HC_8H_4O_4^-?

Explanation / Answer

1)

we know that

moles = molarity x volume (L)

so

moles of KHP titrated = 0.052 x 50 x 10-3 = 0.0026

so

0.026 moles of kHP were titrated

2)

now

from the net ionic equation

we can see that

at equivalence point

moles of KHP = moles of NaOH added

so

M1V1 = M2V2

so

0.052 x 50 = 0.1 x V2

V2 = 26

so

26 ml of NaOH are needed to reach equivalence point

3)

we can see that

moles of conjugate base = moles of KHP reacted

so

moles of conjugate base = 0.0026

now

final volume = 50 + 26 = 76 ml

now

conc = 0.0026 x 1000 / 76

conc = 0.03421

so

the concentration of conjugate base is 0.03421 M

4)

given

pH = 8.96

pOH = 14 - 8.96

pOH = 5.04

pOH = -log [OH-]

5.04 = -log [OH-]

[OH-] = 9.12 x 10-6

now

the conjugate base is weak

for weak bases

[OH-] = sqrt ( Kb x C)

so

9.12 x 10-6 = sqrt ( Kb x 0.03421)

Kb = 2.43 x 10-9

5)

now

Ka = 10-14 / Kb

so

Ka = 10-14 / 2.43 x 10-9

Ka = 4.11 x 10-6

so

Ka value is 4.11 x 10-6

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