Question

0.4108gram sample of CaCO3 is added to a flask along with15 ml of 2 M HCl Rnough water is then added to make 250 ml of solution .A 20 ml aliquot of solution A is taken and titrated with 0.1160M NaOH How many ml of NaOH are used?

Answer 1

We assess the reaction...`CaCO_3(s) + 2HCl(aq) rarr CaCl_2(aq) + H_2O(l) + CO_2(g)uarr`...we get a volume of approx. `15*mL`..

Explanation:

And so now we address the molar quantities of each reagent...

`n_"calcium carbonate"=(0.4108*g)/(100.09*g*mol^-1)=4.10*mmol`

`n_"hydrochloric acid"=15xx10^-3*Lxx2.0*mol*L^-1=30.0*mmol`..

Now CLEARLY `HCl` is the reagent in excess...and we gots a molar quantity of....

`30.0*mmol-(2xx4.10*mmol)=21.8*mmol`...with respect to `HCl` INXS (that sounds like a good name for a pop band).

And this molar quantity is diluted to a `250*mL` with fresh solvent. And so the new concentration is...

`(21.8xx10^-3*mol)/(250xx10^-3*L)=0.0872*mol*L^-1` with respect to `HCl(aq)`

And we take a `20*mL` aliquot of this solution....and titrate it with `0.1160*mol*L^-1` `NaOH(aq)`

`"volume"_"NaOH"=(20xx10^-3*Lxx0.0872*mol*L^-1)/(0.1160*mol*L^-1)=0.015*L-=15.0*mL`

...there is a lot of arithmetic here....the basic expression we use is...

`"concentration"="moles of solute"/"volume of solution"`...

And we manipulate this quotient appropriately.