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A NaOH solution is standardized against benzoic acid. if a 385.2 mg sample of benzoic acid requires 33.96 ml of NaOH to reach the end point, what is the concentration of the NaOH ?
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if we assume that you dilute that 385,2 mg of benzoic acid in 100 ml of water and then titrate with NaOH and spend 33,96 ml of NaOH then i got that concentration is ~ 0,1 M (0,0928...) first you calculate number of moles of benzoic acid n=m/M then you calculate concentration of benzoic acid with asumption that you dilute it in 100 ml c=n/V then you use c1V1 = c2V2 and thats it...
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