2. An approximately 0.1 M NaOH solution was standardized with KHP by titration t

Question

2. An approximately 0.1 M NaOH solution was standardized with KHP by titration to a phenolphthalein endpoint. From the following data, calculate the average molarity of the NaOH. The molar mass of KHP is 204.22 g/mol. Trial 1 Trial 2 Mass of weighing paper + KHP 2.1467 g 1.4125 g 2.1719 g 1.4136 g Mass of weighing paper Mass of KHP Moles of KHP Final buret reading 0.73 0.75 834 31.80 mL 32.67 mL 0.15 mL 0.04 m Initial buret reading Volume of NaOH (mL) Molarity of NaOH (moles per liter) Average molarity 3. Calculate the average molarity, standard deviation (SD), and percent (%) variation of the three titration molarities given below. Recall: % variation = (SD/mean)*(100%). Molarity of HCl: 0.3087 M 0.3083 M 0.3092 M Average Molarity (SD) [UNITS!] Percent (%) variation: Is the percent variation less than 1%? Page 131 CHM 277 General Chemistry Laboratory 2017-2018

Explanation / Answer

Trial 1)

moles of KHP = mass/MW = 0.7342/204.22 = 0.003595 mol

V of NaOH = (Vfinal - Vinitial) = 31.80-0.15 = 31.65 mL

[NAOH] = mol of KHP / Vtotal = 0.003595 / (31.65*10^-3) = 0.11358 M

Trial 2:

moles of KHP = mass/MW = 0.7583/204.22 = 0.0037131 mol

V of NaOH = (Vfinal - Vinitial) = 32.67-0.04= 32.63 mL

[NAOH] = mol of KHP / Vtotal = 0.0037131/ (32.63*10^-3) = 0.11379 M

Average molarity --> = (0.11358+0.11379 )/2 = 0.113685 M

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