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 intial buret reading was 2.30 mL. NaOh was added until the endpoint was reached. The final buret reading was 16.60 mL.

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

Given volume of sample = 25.0 mL .

Volume of NaOH used to react = 16.60 mL = 0.0166 L

a) We have to determine number of moles of OH-

Number of moles of OH = Volume of NaOH in L x molarity of NaOH

= 0.0166 L * 0.04125 M

= 0.000685 mol

b)

Reaction between NaOH and KHP

NaOH (aq) + KHC8H4O4 (aq) -- > KNaC8H4O4 (aq) + H2O (l)

In this reaction the mole ratio between NaOH and HP- is one : one

So number of moles of OH- = number of moles of HP-

Therefore answer = 0.000685 mol HP-

c)

We now have the moles of HP- in that we can get the concentration of HP- in 25 mL = 0.025 L of its original.

[HP-]= 0.000685 mol / 0.025 L

=0.0274 M

So the concentration HP- = .00274 M

d)

solubility product constant = [HP-][K+] = x^2 = 0.0274 M x 0.0274 M

x (solublity of KHP ) = 0.0274 M

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