Question

A $50 mL$ sample of hard water containing $Ca^(2+)$ and $Mg^(2+)$ ions is titrated with $50 mL. 0.005 M EDTA$ solution at $pH=10$, using eriochrome balck$-T$ indicator to reach equivalence point.
In a equal amount of hard water sample, $Mg^(2+)$ ions are precipitated as $Mg(OH)(2)$ by adding suitable amount of $NaOH$. the solution, after precipitation of $Mg(OH)(2)$, is stirred and then titrated with $EDTA$ solution using calcon as indicator, and it requires $10 mL$ of above $EDTA$ solution to reach equivalence point.
a. Calculate the strength of $Ca^(2+)$ and $Mg^(2+)$ ions present in hard water.
b. Calculate the hardness due to $Ca^(2+)$ ions in p p m of $CaCO(3)$.
c. Calculate the hardness due to $Mg^(2+)$ ions in p p m of $CaCO(3)$.
d. Calculate the total hardness of water in p p m of $CaCO_(3)$.

Answer 1

Case I (using eriochrome *****-T indicator)
$M_1V_1(EDTA)=M_2V_2(Ca^(2+) and Mg^(2+)$ in $H_2O)$
$50xx0.005=M_2xx50$
$M_2 of Ca^(2+) and Mg^(2+)=0.005M=5m mol L^(-1)$
Case II (Using calcon indicator)
$M_1V_1(EDTA)=M_2V_2(Ca^(2+)$ in $H_2O$)
$10xx0.005=M_2xx50$
$M_2 of Ca^(2+)=10^(-3)M=1.0m mol L^(-1)$
$mmol L^(-1) of Mg^(2+)=5-1=4m mol L^(-1)$
Strength of $Mg^(2+)=4xx24xx10^(-3)=0.096g L^(-1)$
Strength of $Ca^(2+)=1xx40xx10^(-3)=0.04 g L^(-1)$
(b). Hardness due to $Ca^(2+)$ ions of the sample in gram of $CaCO_3$ in $10^(6)$ " mL of " $H_2O$
$=(M(Ca^(2+))xxMw(CaCO_3)xx10^(6))/(10^(3))$
$=(10^(-3)xx100xx10^(6))/(10^(3))=100ppm$
(c). Hardness due to $Mg^(2+)$ ions of the sample in g of $CaCO_3$
in $10^(6)$ " mL of " $H_2O=(MMg^(2+)xxMw(CaCO_3)xx10^(6))/(10^(3))$
(d). Total hardness $=(M(Ca^(2+) and Mg^(2+))xxMw(CaCO_3)xx10^(6))/(10^(3))$
$=(5xx10^(-3)xx100xx10^(6))/(10^(3))=500ppm$
(Alternatively, Total hardness$=$ Hardness due to $Ca^(2+)+$ Hardness to $Mg^(2+)=100+400=500ppm)$