(8 points) The solubility of the weak acid potassium hydrogen phthalate (KHP, 20

Question

(8 points) The solubility of the weak acid potassium hydrogen phthalate (KHP, 204.22 g/mol, pKA= 5.4) s 80.00 g/L in pure water at 293K. You have a 50 mole tablet of KHP to which you add to 10.0L of pure H20 and allow the system to become fully saturated. What is the concentration of KP (the conjugate base of KHP) at equilibrium for the saturated system? Report your final concentration with 3 significant digits. Utilize the ICE Table Method and show your work to obtain any credit. For this question we will assume the activity coefficients for all species to be equal to 1.0. You can assume the volume of the entire solution is also 10.OL.

Explanation / Answer

Molar mass of KHP = 204.22 g/mol

Moles of KHP in a tablet = 50 mol

Mass of KHP = molar mass*moles = 204.22*50 = 10211 g

Total volume of water = 10.0 L

solubility = 80.00 g/L

So, mass of KHP in 10.0 L water = 10.0*80.00 = 800.00 g

So, saturated system contains 800.00 g of KHP in 10.0 L

concentration of KHP = mass/(molar mass*volume) = 800.00/(204.22*10.0) = 0.392 M

KHP + H2O ----> KP- + H3O+

ICE table:

pKa = 5.4

Ka = 10-5.4 = 4.0*10-6

Ka = [KP-][H3O+]/[KHP]

4.0*10-6 = x.x/(0.392-x)

4.0*10-6 = x.x/(0.392)

x2 = 1.56*10-6

x = 1.25*10-3

At equilibrium, [KP-] = x = 1.25*10-3 M

[KHP] [KP-] [H3O+] Initial 0.392 M 0 0 Change -x +x +x Equilibrium 0.392-x x x

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