Jack makes a solution by adding $2\mathrm{.}5\backslash$times $106$ moles of cadmium $($Cd$2+)$ and $1\backslash$times $103$

moles of EDTA to $1$ L of water, waits $5$ minutes, and then drinks it$.$ He claims that he

if waits that long, the complex concentrations reach equilibrium values, and under

these conditions, the remaining free Cd$2+$ is nontoxic. You take your pH probe and

determine that $[$H$+]=107\mathrm{.}5$ in the solution. Given the following information,

determine the free $[$Cd$2+]$ in the solution that Jack drinks. You may ignore the

protonated forms of EDTA$4.$ What is the concentration of free Cd$2+$ when the

solution reaches his stomach $($pH $=1)?$

Cd$2++$ H$2$O $=$ CdOH$++$ H$+$ Cd$2++2$H$2$O $=$ CdOH$2+2$H$+$ Cd$2++3$H$2$O $=$ CdOH$3-+3$H$+$ Cd$2++4$H$2$O $=$ CdOH$42-+4$H$+$ Cd$2++$ EDTA$4-=$ Cd$-$EDTA$2-$

$*$b$1=10-10\mathrm{.}08*$b$2=10-20\mathrm{.}35*$b$3=10-33\mathrm{.}30*$b$4=10-47\mathrm{.}35$

b$1=103\mathrm{.}9$