Question

Zinic Chloride in a 50.0ml sample of wateris titrated with 1.87 X 10^-5 M EDTA solution, requiring 24.70ml of the complexing agent. What is the zinic concentration in the water sample expressed as;

a) mol/L?

b) mmol/L

c) g/L

d) ppm

with strach

Answer 1

a) 9.24×10^-6 mol/L

b) 9.24 × 10^-3 mmol/L

c) 6.04 ×10^-4 g/L

d) 0.604 ppm

Explanation

Reaction between EDTA and Zn²⁺ is 1:1 molar reaction

molarity = number of moles per liter of solution

Number of moles of EDTA consumed = (1.87×10^-5mol/1000ml)× 24.70ml = 4.6189×10^-7mol

4.6189×10^-7mol of EDTA react with 4.6189×10^-7 mol of Zn²⁺

Therefore

Number of moles of Zn²⁺ present in the 50.0ml of water = 4.6189×10^-7mol

Concentration of Zn in water sample in mol/L = ( 4.6189×10^-7mol/50.0ml)×1000ml = 9.24×10^-6mol/L

Concentration of Zn in water sample in mmol/L = 9.24×10^-6 mol/L × 1000 = 9.24×10^-3 mmol/L

Molar mass of Zn = 65.38g/mol × 9.2378×10^-6mol = 6.04×10^-4 g = 0.604mg

Concentration of Znin water sample in g/L = 6.04×10^-4g/L

Concentration of Zn in water sample in ppm = 0.604mg/L = 0.604ppm