Question

Name Section/CRN EXPERIMENT 9 POTENTIOMETRIC DETERMINATION OF AN EQUILIBRIUM CONSTANT PRE-LABORATORY QUESTIONS The following preparatory questions should be answered before coming to lab. They are intended to introduce you to several jdeas that are important to aspects of the experiment. You must turn in your work to your instructor before you will be allowed to begin the experiment. Potassium acid phthalate, KHP (KHC4H&O4), is a primary standard reagent used to determine exactly the concentration of a solution of base, such as NaOH, whose approximate concentration is known. KHP is a weak acid, and the equation for the neutralization of KHP by NaOH is Potassium sodium phthalate .COOK COOK + Н.о NaOH COONa сОон КНР Potassium hydrogen phthalate 1. A Chem1515 student dissolved 0.33 grams of KHP in 25.0 mL of water. He found out that it required 18.5 mL of NaOH to reach the endpoint of the titration. a. How many moles of the acid, KHP, were originally present before the addition of the base? b. How many moles of NaOH were required to neutralize the amount of KHP in Part a?

Answer 1

11 mass KH P = 0.33g Volume water = 25mL molar mass KHP = 204.22 g/mol Molarity = mass molar mass X Volume (L) = 0.33 g 204:22g x 25 x103 L MOL M=0:0646M?

a) moles KHP= mass Molar mass = 0.338 204.22 g/mol = 11.616*10-3 mol.] (6) : KHP & Nach reacts in 1:1 mole ratio Moles of Nash required = 11.616X163 mol. I c) Conc. = moles Volume (L)

= 1.616xlo3 mol 18.5x10-3 L NaOH = 0.08735M] Cook not look Cool (Base) el coona (Salt) Lol + Nach To + H₂O (acid) (Base) At end borind call base reacted with and to form salt. So, no aud-base remaining Here; 1.616x16 3 mol of Allach reacted with 1.616 Xlo 3 ml. of KHP to produce 1.6168103 mol of salt total volume = 25ml +18.5mL = 43.5 m2 = 43.5x10 L Salt] = moles = 1.616X163 mol Volume 43.5x10-3 L = 0.03715M / (CHP) Salt is of weak Base acid and Strong base (NaOH). :: PH= 7+ - 2 [pka + loge C: conc. of at salt Ka= 3.89 xlob pka = -log ka = 5.41 pH = 7+ 1 [5.41 +log (0.03715) pH=8.99 Ther, PH at end point was found to be 8.99