Answer to Question #216561 in General Chemistry for Mei

Question

Titration of Ca2+ and Mg2+ in a 50.00-mL sample of hard water required 23.65 mL of 

0.01205 M EDTA. A second 50.00-mL aliquot was made strongly basic with NaOH to precipitate 

Mg2+ as Mg(OH)2. The supernatant liquid was titrated with 14.53 mL of the EDTA solution. 

Calculate

a) The total hardness of the water sample, expressed as ppm CaCO3

b) The concentration of CaCO3 and MgCO3 in the sample in ppm

Answer 1

Solution

Volume used= 23.65ml

$23.65ml EDTA\times\frac{0.01205mmolEDTA}{molEDTA} = 0.2850 mmol$

1 Mol EDTA= 1 Mol Ca²⁺+ Mg²⁺

(0.2850mlCa²⁺).(0.05L).($\frac{1molEDTA}{1molCa2^{2+}})$

=0.01425 moles

Concentration of CaC0₃ and MgC0₃

Molarity of EDTA used

=$\frac{0.01425}{0.01453}= 0.901$M

$(\frac{0.901molEDTA}{L}).( \frac{0.01453L}{1})(\frac{1 molEDTA}{1molca_{2+}})$

= 0.01309 moles