**Re: Chemistry Titration Word Problem. Please Help?** Using the percentage you have for KHP:
Each 100 gm of the crude has 85 gm of KHP and accordingly if you have 2.5 gm of the crude sample, you can calculate the exact amount of KHP as follows:
(2.5*85)/100 = 2.125 gm of KHP.
By the way KHP is monoacidic, and what that means is KHP can react with NaOH (monobasic) in reaction ratio of one to one. To solve this problem, we need to know the molecular weight of KHP which is 204.2212 gm/mol. Accordingly, the no. of moles of KHP you have = Weight/Molecular weight = 2.125/204.2212 = 0.0104 mol.
As the reaction with NaOH is one to one, in titration we will need same no of moles of KHP to get the neutralization point. So we will need 0.0104 moles of NaOH.
As a result by knowing the Molarity and the moles of NaOH, we should be able to determine how much volume of NaOH will be required for 2.5 gm of 85%KHP, as follows:
Volume (L) = Moles/Molarity = 0.0104/0.402= 0.02588 L = 25.88 mL.... |