Question

Calculate the concentration of H₂Y^2- at the equivalence point in Exercise 12-C

Short answer

H₂Y^2- concentration at equivalence point will be 1.1×10^-7M

Step-by-step solution

Given Information

With reference to 12-c the following information were given

Amount of EDTA required for titration = 50 mL of 0.04 M EDTA

Concentration of Cu(NO₃)₂ required for titration =0.08 M

pH maintained =5

EDTA

EDTA (Ethylenediaminetetraacetic acid) is one type of chelating agent. It is used to treat heavy metal toxicity, neurotoxicity, lead poisoning, coronary artery disease etc. It is also used in different industries like pulp and paper, textile etc. It has applications in biochemistry and molecular biology lab.

Determine the concentration

From 12-C the equivalence point was determined at 1.24×10^-7M EDTA concentration

$$\begin{gathered} K=K_5{K}_6=\frac{{\left[H^{+}\right]}^2\left[Y^{4-}\right]}{\left[H_2{Y}^{2-}\right]} \\ \left[H_2{Y}^{2-}\right]=\frac{{\left[H^{+}\right]}^2\left[Y^{4-}\right]}{K_5{K}_6} \\ \left[H_2{Y}^{2-}\right]=\frac{{\left[H^{+}\right]}^2{\alpha }_{Y^{4-}}\left[EDTA\right]}{K_5{K}_6} \\ \left[H_2{Y}^{2-}\right]=\frac{{\left[{10}^{-5}\right]}^2\left[2.9\times {10}^{-7}\right]\left[1.24\times {10}^{-7}\right]}{K_5{K}_6} \\ \left[H_2{Y}^{2-}\right]=1.1\times {10}^{-7}M \end{gathered}$$

Therefore, H₂Y^2- concentration at equivalence point will be 1.1×10^-7M