A student was given 25.0 mL sample of a saturated salt solutions, KHP. it was di

Question

A student was given 25.0 mL sample of a saturated salt solutions, KHP. it was diluted in 25 mL of DI water and a drop of indicator was added. (note that HP- does not dissociate to any appreciable extent when dissolved in water.) it was tirated with 0.04125 M NaOH. the inital reading was 2.30mL. NaOH was added until the endpoint was reached. The final buret reading was 16.60mL.

a. determine the number of moles of OH- used in the titration

b. determine the number of moles of HP- in the original solution

c calculate the concentration of the orignal saturated solution of KHP

d. calculate the solubility product constant of the salt, KHP

Explanation / Answer

a )

volume of OH- used = 16.60 - 2.30 = 14.30 mL

moles of OH - = molarity x volume / 1000

= 0.04152 x 14.30 / 1000

= 5.89 x 10^-4

b)

HP - + OH- ----------------------> P- + H2O

moles of OH- = moles of HP-

so HP- moles = 5.89 x 10^-4

c)

molarity = moles / volume

= 5.89 x 10^-4 / 25 x 10^-3

= 0.02356 M

d)

[K+] =[HP-] = 0.02356 M

Ksp = [K+] [HP-]

Ksp = 5.55 x 10^-4

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