Question

Potassium ion in a 250.0 (±0.1) mL water sample was precipitated with sodium tetraphenylborate:

${\mathit{K}}^{\mathbf{+}}\mathbf{+}{\left(C_6{H}_5\right)}_{\mathbf{4}}{\mathit{B}}^{\mathbf{-}}\mathbf{\to }\mathit{K}\mathit{B}{\left(C_6{H}_5\right)}_{\mathbf{4}}\left(s\right)$

The precipitate was filtered, washed, dissolved in an organic solvent, and treated with excess Hg (EDTA)^2-:

$\mathbf{4}\mathit{H}\mathit{g}{\mathit{Y}}^{\mathbf{2}\mathbf{-}}\mathbf{+}{\left(C_6{H}_5\right)}_{\mathbf{4}}{\mathit{B}}^{\mathbf{-}}\mathbf{+}\mathbf{4}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\to }{\mathit{H}}_{\mathbf{3}}\mathit{B}{\mathit{O}}_{\mathbf{3}}\mathbf{+}\mathbf{4}{\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{5}}\mathit{H}{\mathit{g}}^{\mathbf{+}}\mathbf{+}\mathbf{4}\mathit{H}{\mathit{Y}}^{\mathbf{3}\mathbf{-}}\mathbf{+}\mathit{O}{\mathit{H}}^{\mathbf{-}}$

The liberated EDTA was titrated with 28.73 (±0.03) mL of 0.043 7 (±0.000 1) M Zn²⁺. Find [K⁺] (and its absolute uncertainty) in the original sample.

Short answer

The amount of K⁺ in the original sample will be 1.256(±0.003) mM

Step-by-step solution

Given information

Amount of water sample = 250 mL

For each mole of K⁺ involved in the first reaction , 4 moles of EDTA generated in the second reaction .

Amount of Zn²⁺ used for EDTA titration=28.73 (±0.03) mL of 0.043 7 (±0.000 1) M

EDTA

EDTA also known as Ethylenediaminetetraacetic acid is one type of chelating agent used in different industries, medications and laboratory applications. It is generally used to treat heavy metal toxicity, lead poisoning, coronary artery disease, neurotoxicity etc.

Determine K+

$$\begin{gathered} \left[K^{+}\right]=\frac{{\displaystyle \frac{1}{4}}\left(molesofZ{n}^{2+}\right)}{Volumeofsample} \\ =\frac{{\displaystyle \frac{1}{4}}\left[\left(28.73\right)\left(\pm 0.03mL\right)\right]\left[0.0437\left(\pm 0.0001M\right)\right]}{250\left(\pm 0.1\right)} \\ =\frac{{\displaystyle \left[\frac{1}{4}\left(\pm 0\%\right)\right]\left[\left(28.73\right)\left(\pm 0.104\%\right)\right]\left[0.0437\left(\pm 0.229\%\right)\right]}}{250\left(\pm 0.040\%\right)} \\ =1.256\left(\pm 0.265\right)\times {10}^{-3}M \\ =1.256\left(\pm 0.003\right)mM \end{gathered}$$

Therefore, the amount of K⁺ in the original sample will be 1.256(±0.003) mM.