Question

A mixture having a volume of $\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mL}$and containing $\mathbf{0}\mathbf{.}\mathbf{1000}{\mathbf{MAg}}^{\mathbf{+}}$and${\mathbf{\text{0.1000MHg}}}_{\mathbf{2}}^{\mathbf{2}\mathbf{+}}$was titrated with $\mathbf{0}\mathbf{.}\mathbf{1000}\mathbf{MKCN}$to precipitate ${\mathbf{Hg}}_{\mathbf{2}}{\left(\mathrm{CN}\right)}_{\mathbf{2}}$and $\mathbf{AgCN}$.

(a) Calculate ${\mathbf{pCN}}^{\mathbf{-}}$at each of the following volumes of added KCN: $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{15}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{19}\mathbf{.}\mathbf{90}\mathbf{,}\mathbf{20}\mathbf{.}\mathbf{10}\mathbf{,}\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{30}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{35}\mathbf{.}\mathbf{00}\mathbf{mL}$

(b) Should any AgCN be precipitated at $\mathbf{19}\mathbf{.}\mathbf{90}\mathbf{mL}$?

Using Spreadsheets

Short answer

a) The ${\mathrm{pCN}}^{-}$at $5.00,10.00,15.00,19.90,20.10,25.00,30.00,35.00\mathrm{mL}$of volumes are $18.85,19.00,18.65,17.76,14.17,1.845,1.60,1.48$.

b) Because the equivalency point is $20\mathrm{mL}$and the increased volume of potassium cyanide is $19.90\mathrm{mL}$, no silver cyanide precipitates. As a result, no precipitation occurs

Step-by-step solution

Concept used.

The ${\mathrm{pAg}}^{*}$values of silver ions of given titration has to be calculated.

Concept introduction:

$\mathrm{pX}=-{\log }_{10}\left[\mathrm{X}\right]$

p$\left[\mathrm{X}\right]$is concentration of$\mathrm{X}$

The ratio of moles of solute (in grammes) to the volume of the solution is known as molarity (in litres). The formula for calculating a solution's molarity is:

$\mathrm{Molarity}\left(\mathrm{M}\right)=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}\left(\mathrm{ing}\right)}{\mathrm{volume}\mathrm{of}\mathrm{solution}\left(\mathrm{inL}\right)}$

Calculate the pAg+values of silver ions of 10.00mLtitration has to be calculated.

a)

Because mercuric cyanide's Ksp is lower than silver cyanide's, the first equivalency point where mercuric cyanide precipitates is computed as

The ratio of moles of cyanide ion to moles of mercuric ions is $2:1$

$$\begin{gathered} {\mathrm{v}}_{\mathrm{e}}\times 0.1000\mathrm{M}=0.010\mathrm{L}\times 0.1000\mathrm{M} \\ {\mathrm{v}}_{\mathrm{e}}=\frac{0.010\mathrm{L}\times 0.1000\mathrm{M}}{0.1000\mathrm{M}} \\ =0.010\mathrm{L} \\ =10\mathrm{mL} \end{gathered}$$

At the equivalency point, the concentrations of mercuric and cyanide ions are the same .

$$\begin{gathered} \left[{\mathrm{Hg}}_2^{2+}\right]{\left[{\mathrm{CN}}^{-}\right]}^2=4{\mathrm{x}}^3 \\ ={\mathrm{K}}_{\mathrm{sp}}(\mathrm{for}{\mathrm{Hg}}_2{\left(\mathrm{CN}\right)}_2) \\ \left[{\mathrm{Hg}}_2^{2+}\right]{\left[{\mathrm{CN}}^{-}\right]}^2=4{\mathrm{x}}^3 \\ =5.0\times {10}^{-40}\left[{\mathrm{CN}}^{-}\right] \\ =1.113\times {10}^{-19}\mathrm{M} \\ {\mathrm{pCN}}^{-}=18.85 \end{gathered}$$

When $5\mathrm{mL}$of cyanide ion is added, the concentration of cyanide ion is

$$\begin{gathered} {\mathrm{CN}}_2^{-1}=\frac{{\mathrm{K}}_{2\mathrm{p}}}{\left[{{\mathrm{H}}_{{\mathrm{g}}_2}}^{2+}\right]} \\ =\frac{5.0\times {10}^{40}}{\left({\displaystyle \frac{10\mathrm{mL}-5\mathrm{mL}}{10\mathrm{mL}}}\right)\left(0.1000\mathrm{M}\right)\left({\displaystyle \frac{1000\mathrm{ml}.}{15.00\mathrm{ml}.}}\right)} \\ =1.225\times {10}^{-19}\mathrm{M} \\ {\mathrm{pCN}}^{*}=-\log \left[{\mathrm{CN}}^{-1}\right] \\ =-\log \left[1.225\times {10}^{-19}\right] \\ =19.00 \end{gathered}$$

When$15\mathrm{mL}$of cyanide ion is added, the concentration of cyanide ion is

$$\begin{gathered} \left[{\mathrm{CN}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{{\mathrm{Ag}}^{+}} \\ =\frac{2.2\times {10}^{-16}}{\left({\displaystyle \frac{20\mathrm{mL}.13.00\mathrm{ml}.}{10\mathrm{mL}}}\right)\left(0.1000\mathrm{M}\left({\displaystyle \frac{10.00\mathrm{l}.}{2.00\mathrm{mL}.}}\right)\right)} \\ =2.24\times {10}^{-19}\mathrm{M} \\ {\mathrm{pCN}}^{-}=-\log \left[{\mathrm{CN}}^{-}\right] \\ =-\log \left[2.24\times {10}^{-19}\right] \\ =18.65 \end{gathered}$$

When $19.90\mathrm{mL}$of cyanide ion is added, the concentration of cyanide ion is

$$\begin{gathered} \left[{\mathrm{CN}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{[{\mathrm{Ag}}^{+}l} \\ =\frac{2.2\times {10}^{-16}}{\left({\displaystyle \frac{20\mathrm{mL}.10.90\mathrm{ml}m}{10L}}\right)\left(0.1000\mathrm{M}\right)\left({\displaystyle \frac{10.00m\mathrm{l}.}{29.90\mathrm{mL}.}}\right)} \\ =1.74\times {10}^{-18}\mathrm{M} \\ {\mathrm{pCN}}^{-}=-\log \left[{\mathrm{CN}}^{-}\right] \\ =-\log \left[1.74\times {10}^{-18}\right] \\ =17.76 \end{gathered}$$

When $20.10\mathrm{mL}$of cyanide ion is added, there is only cyanide ions present in excess.

$$\begin{gathered} Volumeofcyanideions \\ =\left(2.10-20.00\right) \\ =0.10m{L}^{-}C{N}^{-} \\ \left[C{N}^{-}\right]=\left(\frac{0.10mL}{30.10mL}\right)\left(0.1000M\right) \\ =2.0\times {10}^{-15}M \\ pC{N}^{-}=-\log \left[C{N}^{-}\right] \\ =-\log \left[3.3\times {10}^{-4}\right] \\ =14.17 \\ Volumeofcyanideions \\ =25.00-20.00=5mLofC{N}^{-} \\ \left[C{N}^{-}\right]=\left(\frac{5mL}{35mL}\right)\left(0.1000M\right) \\ =0.0143M \\ pC{N}^{-}=-\log \left[C{N}^{-}\right] \\ =-\log \left[0.0143\right] \\ =1.845 \\ Volumeofcyanideions=\left(30.00-20.00\right)=10mLofC{N}^{-} \\ \left[C{N}^{-}\right]=\left(\frac{10mL}{40mL}\right)\left(0.1000M\right) \\ =0.0250M \\ pC{N}^{-}=-\log \left[C{N}^{-}\right] \\ =-\log \left[0.250\right] \\ =1.60 \\ Volumeofcyanideions=\left(35.00-20.00\right)=15mLofC{N}^{-} \\ \left[C{N}^{-}\right]=\left(\frac{15mL}{45mL}\right)\left(0.1000M\right) \\ =0.033M \\ pC{N}^{-}=-\log \left[C{N}^{-}\right] \\ =-\log \left[0.033\right] \\ =1.48 \end{gathered}$$

Determine if AgCN precipitate at 19.90 mL 

b)

Because the equivalency point is $20\mathrm{mL}$nd the increased volume of potassium cyanide is $19.90\mathrm{mL}$, no silver cyanide precipitates.

As a result, no precipitation occurs.