Prelab2 - 1 mmol NaOH 1 mmol KHP M NaOH x V NaOH = Wt KHP...

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Experiment 2 Standardization: Weigh 0.7-0.9 g KHP for each titration Volume for each will not be the same Unknown Determination: Grade based only on average Wt% KHP Many different unknowns Use approx. one gram unknown Adjust weight to get V ep = 20-50 mL No pH meters Don't boil water (keep bottle capped) No extra unknown after 6 titrations
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Standardization of NaOH What is molarity of NaOH solution if titration of 0.7671 g of primary standard KHP consumes 28.98 mL of NaOH soln? Fundamental Relationship of Titrimetric Analysis: Moles Titrant = Moles Substance Titrated x S.R. mmol NaOH = mmol KHP x
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Unformatted text preview: 1 mmol NaOH 1 mmol KHP M NaOH x V NaOH = Wt KHP mm KHP x 1 mmol NaOH 1 mmol KHP M x 28.98 mL = 767.1 mg 204.23 mg/mmol x 1 mmol NaOH 1 mmol KHP M = 0.1296 mmol/mL Determination of the Acid Unknown Titration of 1.0380 g of sample consumes 35.32 mL of 0.1296 M NaOH. What is weight percent KHP in unknown? M NaOH x V NaOH = Wt KHP mm KHP x 1 mmol NaOH 1 mmol KHP 0.1296 M x 35.32 mL = Wt KHP 204.23 mg/mmol x 1 mmol NaOH 1 mmol KHP Wt KHP = 934.9 mg Wt% KHP = Wt KHP Sample Wt x 100% = 934.9 mg 1038.0 mg x 100% = 90.07%...
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This note was uploaded on 04/09/2008 for the course CHE 1316 taught by Professor Gipson during the Spring '08 term at Baylor.

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Prelab2 - 1 mmol NaOH 1 mmol KHP M NaOH x V NaOH = Wt KHP...

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