Answer:
Incomplete question: 9.9 mL of 0.01 M EDTA, and 10 mL of buffer with a pH of 4.
The concentration of [HY³⁻] is 1.8x10⁻²⁸mol/L
Explanation:
Given information:
10 mL of 0.01 M VOSO₄ = 0.01 L
9.9 mL of 0.01 M EDTA = 0.0099 L
10 mL of buffer = 0.01 L
pH of 4
Question: Calculate HY³⁻, [HY³⁻] = ?
Moles of EDTA:
[tex]n_{EDTA} =0.01\frac{moles}{L} *0.0099L=9.9x10^{-5} moles[/tex]
The concentration of Y⁴⁺:
At pH = 4 Y⁴⁻ = 3.8x10⁻⁹
[tex][Y^{4+} ]=3.8x10^{-9} *9.9x10^{-5} =3.762x10^{-13}[/tex]mol/L
The concentration of [H₂Y²⁻]:
[tex][H_{2} Y^{2-} ]=Kf*[VO_{2} ^{2+} ][Y^{4+} ]=6.9x10^{-7} *(0.01*0.01)*3.762x10^{-13} =2.6x10^{-23}[/tex]
The reaction:
H₂Y²⁻ → H⁺ + HY³⁻
The concentration of HY³⁻
[tex]6.9x10^{-7} =\frac{[H+][HY^{3-}] }{[H_{2}Y^{2-}] } \\6.9x10^{-7} =\frac{0.1*[HY^{3-}]}{2.6x10^{-23} }[/tex]
Solving for HY³⁻
[HY³⁻] = 1.8x10⁻²⁸mol/L
