Hi. I'm having trouble with this:
In order to standardise a solution of sodium hyrdoxide, a chemists first prepared a solution of ethanedioic acid-2-water, H2C2O4 . 2H2O by dissolving 14.6g of crystals in water and making the solution up to 250 cm3 in a graduated flask. He then pipetted 25cm3 of this solution into a conical flask, added phenolphthalein solution as indicator, and titrated it against the sodium hydroxide solution! 24.1cm3 of the latter were required. Calculate, first, the molar concentratoin of the ethanedioic acid solution, and then that of the sodium hydroxide.
moles H2C2O4 . 2H2: 14.6/ 126 = 0.116 moles
concentration of H2C2O4.2H2O: 0.116 / (25/1000) = 4.64 moles
H2C2O4.2H2O + NaOH --> NaHC2O4 + 3H2O ( 1:1 reaction)
4.64/ (24.1 / 1000) = 192.5 mold dm3 (THIS LOOKS LIKE A VERY HIGH CONCENTRATION)
can someone please check it and tell me where I went wrong and guidance on how to get the right answer please