A 0.7500-g sample of pure KHP (Ewt=204.2g/mol) was dissolved and titrated with 35.0 mL of NaOH. The end point was overst

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answerhappygod
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A 0.7500-g sample of pure KHP (Ewt=204.2g/mol) was dissolved and titrated with 35.0 mL of NaOH. The end point was overst

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A 0 7500 G Sample Of Pure Khp Ewt 204 2g Mol Was Dissolved And Titrated With 35 0 Ml Of Naoh The End Point Was Overst 1
A 0 7500 G Sample Of Pure Khp Ewt 204 2g Mol Was Dissolved And Titrated With 35 0 Ml Of Naoh The End Point Was Overst 1 (28.04 KiB) Viewed 6 times
A 0.7500-g sample of pure KHP (Ewt=204.2g/mol) was dissolved and titrated with 35.0 mL of NaOH. The end point was overstepped and it was necessary to backtitrate with 3.50 ml HCl. Calculate the molarity of NaOH and HCl if the ratio is 30.0 ml NaOH to 25.0 ml HCl.
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