# 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 ?

Dec 17, 2016

$\textsf{1.83 \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

$\textsf{S {n}^{2 +} \rightarrow S {n}^{4 +} + 2 e \text{ } \textcolor{red}{\left(1\right)}}$

$\textsf{C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 e \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O \text{ } \textcolor{red}{\left(2\right)}}$

To get the electrons to balance we multiply sf(color(red)((1)) by 3 and add to $\textsf{\textcolor{red}{\left(2\right)} \Rightarrow}$

$\textsf{C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + \cancel{6 e} + 3 S {n}^{2 +} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O + 3 S {n}^{4 +} + \cancel{6 e}}$

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

$\textsf{{n}_{C r \left(V I\right)} = c \times v = 0.495 \times \frac{4.94}{1000} = 2.4453 \times {10}^{- 3}}$

$\therefore$$\textsf{{n}_{S n \left(I I\right)} = 2.4453 \times {10}^{- 3} \times 3 = 7.3359 \times {10}^{- 3}}$

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{\left[S {n}^{2 +}\right] = \frac{7.3359 \times \cancel{{10}^{- 3}}}{\frac{4.00}{\cancel{1000}}} = 1.83 \textcolor{w h i t e}{x} \text{mol/l}}$