195 PRELAB Unless told otherwise, this is a hard copy prelab. Due in lab at the

Question

195 PRELAB Unless told otherwise, this is a hard copy prelab. Due in lab at the start of the lab session. 1. If a 0.315 g f KHP requires 14.85 mL of NaOH, calculate [NaOH]. 5 pts 2. If the same NaOH solution is now used to titrate 0.152 g unknown diprotic acid and requires 26.21 mL to the endpoint, what is the molar mass of the unknown diprotic acid? 5 pts If the same NaOH solution is now used to titrate 0.152 g unknown triprotic acid and requires 23.75 mL to the endpoint, what is the molar mass of the unknown triprotic acid? 5 pts 3.

Explanation / Answer

Q1

mol of KHP = mass/MW= 0.315/204.3 = 0.00154 mol

mol of KHP = mol of NaOH

mol of NaOH = 0.00154

[NaOH] = mol/V = 0.00154/(14.85*10^-3) = 0.10370 M

Q2

if diprotic acid

mmol of NaOH = MV = 0.10370*26.21 = 2.717977

then

mmol of adiprotic acid = 1/2*mmol of base = 1/2*2.717977 =1.3589 mmol of diprotic acid

MM = mass/mol = 0.152/(1.3589*10^-3) = 111.85 g/mol

Q3

if triprotic

mmol of base = MV = 0.10370*23.75*10^-3 = 0.002462

mmol of acid = 1/3*mmol of base = 1/3*0.002462 =0.0008206

MM = mass/mol = (0.152)/(0.00082) = 185.3 g/mol

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