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Part A : molar concentration of unknown acid = 0.04162 M
Part B : volume of NaOH required = 60.45 mL
Part A :
concentration NaOH = 0.1198 M
volume NaOH = 20.15 mL
moles NaOH = (concentration NaOH) * (volume NaOH)
moles NaOH = (0.1198 M) * (20.15 mL)
moles NaOH = 2.414 mmol
moles acid present = 2.414 mmol
concentration of acid = (moles acid present) / (volume of acid)
concentration of acid = (2.414 mmol) / (58.00 mL)
concentration of acid = 0.04162 M
Part A When titrated with a 0.1198 M solution of sodium hydroxide, a 58.00 mL solution...
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