It says in a volumetric analysis experiment, an acidic aqueous solution of methanol(CH3OH) is titrtated with a solution of potassium dichromate according to the following balanced chemical equation


2 K2Cr2O7+ 8 H2SO4+3 CH3OH yields 2 Cr2(SO4)3+ 11 H2O+ 3 HCOOH+ 2 K2SO4


and the question asks what volume of 0.00197 M K2Cr2O7 is required to titrate 1.20 g of CH3OH dissolved in 45 ml of solution?

2 K2Cr2O7+ 8 H2SO4+3 CH3OH yields 2 Cr2(SO4)3+ 11 H2O+ 3 HCOOH+ 2 K2SO4
From this balanced equation
you can see that
2 moles of K2Cr2O7 reacts with 3 moles of CH3OH
No of moles of CH3OH = mass/ molar mass = 1.20/32 = 0.0375
No of moles of K2Cr2O7 required to react with 0.0375 moles of CH3OH = 0.0375 X 2/3 =0.025
  moles = molarity X volume in L
or Volume = moles/molarity
now you can solve it for V

a 60 ml sample of a weak base is titrated with 0.0444M HCl. At the endpoint, it is found that 13.52 ml of titrant was used. What is the concentration of the weak base?

Quote from: carlosjochoa on October 20, 2009, 10:55:48 PM
a 60 ml sample of a weak base is titrated with 0.0444M HCl. At the endpoint, it is found that 13.52 ml of titrant was used. What is the concentration of the weak base?

No of moles of acid used in titration = M X V = 0.01352 L X 0.044M
                         No of moles of base = no of moles acid
   Concentration =  moles /volume of base in L