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 ?

Answer 1

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

To begin, consider 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 indicates a reaction between 1 mole of Cr(VI) and 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)#

The formula for [Sn^(2+)] is 7.3359xcancel(10^(-3)))/(4.00/cancel(1000)) = 1.83 color(white)(x)"mol/l").

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Answer 2

To find the concentration of the Sn^(2+) solution, we can use the equation:

[Sn^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + Sn^{4+} + 7H_2O]

Using the balanced equation, we can see that the mole ratio between Sn^(2+) and Cr2O7^(2-) is 1:1. Thus, the moles of Sn^(2+) is equal to the moles of Cr2O7^(2-).

Given that the volume of Cr2O7^(2-) solution is 0.495 M and the mean titre is 4.94 ml, we can calculate the moles of Cr2O7^(2-) used in the titration:

[moles; of; Cr_2O_7^{2-} = (0.495; mol/L) \times (4.94 \times 10^{-3} L)]

Since the moles of Sn^(2+) is equal to the moles of Cr2O7^(2-), the concentration of the Sn^(2+) solution is equal to the moles of Sn^(2+) divided by the volume of the Sn^(2+) solution (in liters):

[Concentration; of; Sn^{2+} = \frac{moles; of; Cr_2O_7^{2-}}{volume; of; Sn^{2+}; solution}]

[Concentration; of; Sn^{2+} = \frac{moles; of; Cr_2O_7^{2-}}{4.00 \times 10^{-3} L}]

[Concentration; of; Sn^{2+} = \frac{(0.495; mol/L) \times (4.94 \times 10^{-3} L)}{4.00 \times 10^{-3} L}]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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