A solution of 4.00 ml of $sf(Sn^(2+))$ of unknown concentration is titrated against an acidified solution of 0.495M $sf(Cr_2O_7^(2-)$ . The mean titre is 4.94 ml. What is the concentration of the $sf(Sn^(2+))$ solution ?

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Answer 1 · 2016-12-17T18:07:17

$sf(1.83color(white)(x)"mol/l")$

Start with the two 1/2 equations:

$sf(Sn^(2+)rarrSn^(4+)+2e" "color(red)((1)))$

$sf(Cr_2O_7^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_2O" "color(red)((2)))$

To get the electrons to balance we multiply $sf(color(red)((1))$ by 3 and add to $sf(color(red)((2))rArr)$

$sf(Cr_2O_7^(2-)+14H^(+)+cancel(6e)+3Sn^(2+)rarr2Cr^(3+)+7H_2O+3Sn^(4+)+cancel(6e))$

This tells us that 1 mole of Cr(VI) reacts with 3 moles of Sn(II).

$sf(n_(Cr(VI))=cxxv=0.495xx4.94/1000=2.4453xx10^(-3))$

$:.$ $sf(n_(Sn(II))=2.4453xx10^(-3)xx3=7.3359xx10^(-3))$

$sf(c=n/v)$

$:.$ $sf([Sn^(2+)]=(7.3359xxcancel(10^(-3)))/(4.00/cancel(1000))=1.83color(white)(x)"mol/l")$

A solution of 4.00 ml of sf(Sn^(2+))sf(Sn^(2+)) of unknown concentration is titrated against an acidified solution of 0.495M sf(Cr_2O_7^(2-)sf(Cr_2O_7^(2-). The mean titre is 4.94 ml. What is the concentration of the sf(Sn^(2+))sf(Sn^(2+)) solution ?