How can we calculate volume for this question of titration? (Chemistry )

Question

What volume of 0.2535 M NaOH required to titrate 0.8508 g of KHP to stoichiometric end point?

Answer 1

First you need to write out the balanced equation to determine the mole ratio between C8H5KO4_(aq) and NaOH_(aq)

 

The chemical formula for KHP is C8H5KO4

NaOH + C8H5KO4-->NaC8H4KO4 + H2O

 

Because the ratio between C8H5KO4 and NaOH is one to one you will need the same number of moles of NaOH as KHP to reach the equivalence point. So the steps are grams KHP to moles KHP to mmol KHP to mmol NaOH to ml NaOH as your final answer:

 

 

 0.8508 g KHP(1mol/204.22g)(1000mmol/1mol)(1mmol NaOH/1mmol KHP)(1ml NaOH/0.2535mmol)= 16.43ml

Answer 2

To add to Andre's notes above: It's irrelevant that the KHP solution (which is chemically ambiguous, is it K2HPO4 or is it KH2PO4? -- assuming the former, and converting to KNa2PO4, but calculations would also work equally well for converting KH2PO4 to KNaHPO4) *initially* has fewer than 4.166 mmol present as HPO4(-2) ion. As you start adding the NaOH, and converting it eventually to K3PO4, any bits that were initially present as H2PO4(1-), or PO4(3-), or H3O(1+), or OH(1-), will all "come out in the wash". In short, titration is all about "theoretical, complete" reactions. If you want a corny analogy, it's like walking a dog around the block: you can take a dog from any house, walk once around the block, and return it to the same house. The distance "around the block" (amount of titrant required) doesn't change, regardless of the house (equilibrium-driven side conditions) position selected initially.

Now, that's different than asking about pH values in the solution, since the actual [H3O(1+)] level is affected by the various equilibrium reactions the salt ions nominally present may have undergone. (Or, which house did you just return that dog to?)

Hope this helps!

Answer 3

Molar mass of KHP: 204.22 g/mol, so we have 0.8508 g/(204.22 g/mol) = 4.166 mmol of KHP.

 

To titrate it to the equivalence point, we need an equal amount of NaOH: 4.166 mmol of NaOH.

 

With 0.2535 mmol/mL of NaOH, we need a volume of 4.166 mmol/(0.2535 mmol/mL)=16.43 mL.

 

Note: As a weak acid, KHP will not ionize completely (pK_a=5.4), so depending on the initial concentration of KHP we will have fewer than 4.166 mmol available for neutralization.