Question

Calculate [HY^3-] in a solution prepared by mixing 10.00 mL of 0.010 0 M VOSO₄, 9.90 mL of 0.010 0 M EDTA, and 10.0 mL of buffer with a pH of 4.00

Short answer

The concentration of HY^3- in solution is 4.6×10^-11M

Step-by-step solution

Given Information

Volume of VOSO₄ taken to prepare solution = 10 mL

Concentration of VOSO₄ taken to prepare solution = 0.01M

Volume of EDTA taken to prepare solution = 9.90 mL

Concentration of EDTA taken to prepare solution = 0.01M

Volume of buffer (pH=4) solution taken=10 mL

Determine concentration of VO2+ and VOY2-

$\begin{array}{c}\left[V{O}^{2+}\right]=\left(\frac{0.1}{29.9}\right)\left(0.01M\right)\\ =3.34\times {10}^{-5}M\\ \left[VO{Y}^{2-}\right]=\left(\frac{9.9}{29.9}\right)\left(0.01M\right)\\ =3.31\times {10}^{-3}M\end{array}$

Equations and data obtained in order to proceed for calculation are as follows

$\begin{array}{c}{K}_f={10}^{18.7}\\ AtpH4p{K}_6=10.37\left(for{H}_6{Y}^{2+}\right)\end{array}$

Determine concentration of HY3-

$\begin{array}{c}{K}_f=\frac{\left[VO{Y}^{2-}\right]}{\left[V{O}^{2+}\right]\left[Y^{4-}\right]}\\ \left[Y^{4-}\right]=\frac{\left[VO{Y}^{2-}\right]}{\left[V{O}^{2+}\right]K_f}\\ =\frac{\left[3.31\times {10}^{-3}\right]}{\left[3.34\times {10}^{-5}\right]\times {10}^{18.7}}\\ =1.98\times {10}^{-17}\end{array}$

$\begin{array}{c}{K}_6=\frac{\left[H^{+}\right]\left[Y^{4-}\right]}{\left[H{Y}^{3-}\right]}\\ \left[H{Y}^{3-}\right]=\frac{\left[{10}^{-4}\right]\left[1.98\times {10}^{-17}\right]}{\left[{10}^{-10.37}\right]}\\ =\frac{\left[H^{+}\right]\left[Y^{4-}\right]}{K_6}\\ =4.6\times {10}^{-11}M\end{array}$

Therefore,the concentration of HY^3- in solution is 4.6×10^-11M