Chapter 12, Problem 12.AE

Interpretation Introduction

Interpretation:

The concentration of ${\text{K}}^{\text{+}}$ ion in the original sample should be calculated.

Concept introduction:

Molarity:

Molarity is a term of concentration, the number of moles present in Liter of solution is given by molarity and it is given by solution is given the ratio between mole to liter,

$\text{Molarity =}\frac{\text{Number of mole}}{\text{Volume}\text{(Liter)}}$

Answer to Problem 12.AE

The concentration of ${\text{K}}^{\text{+}}$ ion in the original sample is $\text{1}\text{.256}\text{(±0}\text{.003\%)}\text{×}{\text{10}}^{\text{-13}}\text{M}$

Explanation of Solution

To calculate the concentration of ${\text{K}}^{\text{+}}$ ion in the original sample $\text{EDTA}$

Given,

Volume of sample solution is $\text{250}\text{.0}\text{mL}$

Volume of ${\text{Zn}}^{\text{2+}}$ solution is $\text{28}\text{.73 (±0}\text{.03)}\text{mL}$

Concentration of ${\text{Zn}}^{\text{2+}}$ is $\text{0}\text{.0437 (±0}\text{.0001)}\text{mL}$

The reactions are,

$\begin{array}{l}{\text{K}}^{\text{+}}{\text{+(C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}{\text{B}}^{\text{-}}\stackrel{}{\to }{\text{KB(C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4(s)}}\\ \\ {\text{4HgY}}^{\text{2-}}{\text{+(C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}{\text{B}}^{\text{-}}\text{+}{\text{4H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{BO}}_{\text{3}}{\text{+4C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{Hg}}^{\text{+}}{\text{+4HY}}^{\text{3-}}{\text{+OH}}^{\text{-}}\end{array}$

From the above equation, the one mole of ${\text{K}}^{\text{+}}$ reacting in first reaction and 4 molecule of $\text{EDTA}$ reaction in 2 reaction.

From reaction 2, mole of $\text{EDTA}$ is equal to mole of ${\text{Zn}}^{\text{2+}}$

$\begin{array}{l}{\text{Concentration of K}}^{\text{+}}\text{ =}\frac{\frac{\text{1}}{\text{4}}\text{(mole}\text{of}{\text{Zn}}^{\text{2+}}\text{)}}{\text{Volume}\text{of}\text{orginal}\text{sample}}\\ \text{=}\frac{\frac{\text{1}}{\text{4}}\text{[28}\text{.73 (±0}\text{.03) mL][0}\text{.043 7 (±0}\text{.0001) M]}}{\text{250}\text{.0(±0}\text{.1)}\text{mL}}\\ \text{=}\frac{\text{[}\frac{\text{1}}{\text{4}}\text{±(0\%)][28}\text{.73 (±0}\text{.03) mL][0}\text{.043 7 (±0}\text{.0001) M]}}{\text{250}\text{.0(±0}\text{.040}\text{\%)}\text{mL}}\\ \text{=}\text{1256}\text{(±0}\text{.255\%)}\text{×}{\text{10}}^{\text{-13}}\text{M}\\ \text{=}\text{1}\text{.256}\text{(±0}\text{.003\%)}\text{×}{\text{10}}^{\text{-13}}\text{M}\end{array}$

The given values are plugged in above equation to give the concentration of ${\text{K}}^{\text{+}}$ ion in the original sample.

The concentration of ${\text{K}}^{\text{+}}$ ion in the original sample is $\text{1}\text{.256}\text{(±0}\text{.003\%)}\text{×}{\text{10}}^{\text{-13}}\text{M}$

Conclusion

The concentration of ${\text{K}}^{\text{+}}$ ion in the original sample was calculated.