Potassium acid phthalate, KHP (KHC_8H_4O_4), is a primary standard reagent used

Question

Potassium acid phthalate, KHP (KHC_8H_4O_4), is a primary standard reagent used to determine exactly the concentation of a solution of base, such as NaOH, whose approximate concentration is known. KHP is a weak acid, and the equation for the neurtalization of KHP by NaOH is.....

a)C.A. Isernhagen dissolved .48 grams of KHP in 20.0mL of water. She found out that it required 17.0mL of NaOH to reach the endpoint of the titration.

b)How many moles of the acid, KHP, were originally present before the addition of the base?

c)How many moles of NaOH were required to neutralize the amount of KHP in Part b.

d)What is the concentration of the NaOH solution?

e) What is the ppH of the solution at the endpoint? (K_a for KHP = 3.89*10^-6)

Explanation / Answer

b.)No of moles of KHP=0.48/204.227 =2.35 x10^-3

c.)One mole of KHP reacts with one mole of NaoH

So moles of NAOH=2.35 x10^-3

d.)concentration=(2.35 x10^-3)/0.017 = 0.138 M

e.)Kb=[HP-][OH-]/[P2-]

kb= 1 x10^-14/3.89 x10^-6

=2.6 x10^-9

[P2-]=2.35 x10^-3/(0.017 + 0.02) =0.0635 M

So kb=x*x/(0.0635 -x)=2.6 x)10^-9

As Kb is very small

x=sqrt(2.6 x10^-9 x0.0635)=1.285 x10^-5 M=[OH-]

[H+]=1 x10^-14/1.285 x10^-5

=7.78 x10^-10 M

So pH=-log(7.78 x10^-10)=9.11

For the second part,

moles of KHP=moles of NaOH

0.33/204.227=1.285 x10^-5*Volume of NaOH

So VOl of NaOH=125.75 L=125750 mL

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