Question

The drug Librium gives a polarographic wave with$E_{1/2}=-0.265V$(versus S.C.E.) in 0.05$M{H}_{2}S{O}_4$. A 50.0 - Mlsample containing Librium gave a wave height of$0.37\mu A$. When 2.00mLof 3.00mMLibrium in$0.05M{H}_{2}S{O}_4$were added to the sample, the wave height increased to$0.80\mu A$. Find the molarity of Librium in the unknown.

Short answer

The concentration of Librium in given unknown solution is 0.096mM .

Step-by-step solution

Define Molarity.

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

Molarity$\left(M\right)=\frac{Molesofsolute\left(ing\right)}{Volumeofsolution\left(inL\right)}$

 Determine the Molarity of Librium.

Given,

$0.05M{H}_{2}S{O}_4$

$l_x=0.37\mu A$(wave height)

$l_s=0.80\mu A$(wave height after addition of sulfuric acid)

$$\begin{gathered} \frac{{\left[X\right]}_i}{{\left[S\right]}_f+{\left[X\right]}_f}=\frac{l_x}{l_s+X} \\ \frac{x\left(mM\right)}{3.0mM\left({\displaystyle \frac{2.00mL}{52.00mL}}\right)+xmM\left({\displaystyle \frac{50.00}{52.00}}\right)}=\frac{0.37\mu A}{0.80\mu A} \\ \frac{x}{0.1154mM+0.9615x\left(mM\right)}=0.4625 \\ x=0.4625\times \left(0.1154mM+0.9615x\left(mM\right)\right) \\ 0.553x=0.0534 \\ x=0.096mM \end{gathered}$$

The concentration of Librium in given unknown solution is 0.096mM .