Titration calculations
Ok so 10cm3 of HCL of unknown concentration was pipettes into a 250cm3 volumetric flask and made up to the mark. 25cm3 of 0.150 moldm-3 NaOH was pipetted into a conical flask.It was found by titration that a mean volume 22.4cm3 of the diluted HCL was required to neutralise this.
(a) calculate moles in NaOH
(b) hence calculate the moles in HCL in the mean titre.
(c) hence calculate the moles of HCL in the whole 250cm3
(d) calculate the concentration of the original HCL solution.
could someone tell me the answers to each part to check if I have done them correctly ??
Thanks !!!
(a) calculate moles in NaOH
(b) hence calculate the moles in HCL in the mean titre.
(c) hence calculate the moles of HCL in the whole 250cm3
(d) calculate the concentration of the original HCL solution.
could someone tell me the answers to each part to check if I have done them correctly ??
Thanks !!!
Original post by AlphaNick
hmmm how about you post your answers instead because thats more orthodox
For (a) I got 3.75 x10-3
(B) same as 1:1 ratio
(C) 0.04185mol
(D) 4.19 moldm-3
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