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OpenStudy (anonymous):
How many mL of 2.155 M KOH are required to titrate 25.00 ml of 0.3057 M CH3CH2COOH ?
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OpenStudy (callisto):
no of mole of CH3CH2COOH = 0.025x0.3057 = 7.6425x10^3 no of mole of KOH = no of mole of CH3CH2COOH =7.6425x10^3 vol. of KOH required = 7.6425x10^3/2.155 x 1000 = 3.55mL hopefully i don't get it wrong :(
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