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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.
- Spring '08