Question

A 1000 ml solution contains 5.55 g of CaCl₂ and 4.77 g of MgCl₂ .

a)How many mililiters of 0.02 N standard versenate solution would be required in the API titration for total hardness?(Answer:10 ml per 1 ml sample)

b)Express the concentration of Ca⁺² and Mg⁺² in parts per million.(Answer: 5550 and 4770)(approximate for p_s=p_w)

Answer 1

Solution :-

part a) Lets calculate the moles of CaCl2 and MgCl2

Moles = mass / molar mass

Moles of CaCl2= 5.55 g/110.98 g per mol = 0.050 mol

Moles of MgCl2 = 4.77 g / 95.21 g per mol = 0.0501 mol

Total moles of Ca^2+ and Mg^2+ = 0.0500 mol + 0.0501 mol = 0.1001 mol

Volume of solution is 1000 ml that is 1 L

Therefore the total mole concentration = 0.1001 mol /1 L = 0.1001 M

1 Molar = 2 Normal

Because Mg^2+ and Ca^2+ are having 2 valency

Therefore the normal concentration would be

0.1001 M * 2 N / 1 M = 0.2002 N

Now lets calculate the volume of the versenate

0.2002 N * /0.02 N = 10

Hence we need 10 ml versenate for 1 ml of the CaCl2 and MgCl2 solution.

Parr b)

Part per million (ppm) means mg substance per liter or per kg solution

Therefore lets first convert the masses from the gram to mg

5.55 g CaCl2 * 1000 mg / 1 g = 5550 mg CaCl2

4.77 g MgCl2 * 1000 mg / 1 g = 4770 mg MgCl 2

Ppm CaCl2 = 5550 mg / 1 L = 5550 ppm CaCl2

Ppm MgCl2 = 4770 mg / 1 L = 4770 ppm MgCl2