Question

SAMPLE CALCULATIONS A standardized solution of 0.1024 M sodium hydroxide (NaOH) is used to titrate an unknown mixture containing the weak acid potassium acid phthalate (KHP, 204.22 g/mol). The following data was collected. Use this data to determine the weight percent of KHP in the unknown sample. TRIAL 1 TRIAL 2 Mass of Unknown Acid (KHD 0.972 1.098 Initial buret reading 5.75 mL mL 13.63 Final buret reading 30.15 41.12 mL mL Volume NaOH 24.4 mL Moles of NaOH moles moles .005311 moles Moles of KHP O.ppy7 moles Grams of KHP in mixture % Weight% KHP in mixture % % Average weight% final reading nitia = 30.16-5.75 Olume NaoH T2= 41.12O13-63 = 21.49ML TI au.4mL
i have a chemistry lab practical tomorrow. These are the sample calculations to prepare us for the practical. Please give the answers for each trial and HOW you got them so that I know what to do tomorrow

Answer 1

Trial-1 mass unknown = 0.9729 Volume Naok= 24.4mL Molarity NaoH = 0.1024M moles NaoH= molarity & Volume (2) = 0.1024 mol % 24.4x16 -3K

= 2.49856 x 10-3 md Reaction Loona o + Nach coth,0 KHP Nakp " Base and KHP rearts in lil mole ratio. Moles KHP = 2.49856x10 mot → gram KHP = molar mass x moles = 204-224 x 2.49956 x 63ml mol = 0.510256 g Weight & KHP = gram KHP xloo gram mixture = 0.5102568 xlor 0.9729 = 152.5% Trial-2 Mais unknown = 1.0989 Volume NaOH = 27.49mL Molarity NaOH = 0.1024M moles NAOH = 0.1024 mol x 27.49 xlos = 2.815 xlo 3 no Moles KHP = moles Nool1 = 2.615 xto 3 mol gram KHP = 204.229 x 2.815xl 3 mb pol = 0.57499 weight!. KHp = 0.57498 xloo 7.098g - 52.36). Average oti els = 52.54, +52.2 = 5204 341. This, average til was found to be 52.438 2

Please correct your moles of KHP. I have mentioned it in red box