# Hardness and Alkalinity Numerical | Water Supply Engineering

### 1. Formula

 Alkalinity due to B– as CaCo3 = alkalinity as B– x eq. wt. of CaCo3 / eq. wt. of B– Hardness due to M++ as CaCo3 = hardness as M++ (mg/lit) x eq. wt. of CaCo3 / eq. wt. of M++ Total Hardness = Carbonate hardness + Non carbonate hardness [H+] = antilog10 [-pH] ,

If, TH > A

then, CH = A

NCH = TH-CH

NCH = TH-A

If, TH =<A

then, CH= TH

NCH=0

### 2. Hardness and Alkalinity Numerical

1. The analysis of water from a well showed the following results in mg/lit;

Ca= 65, Mg=51, Na=101.5, K= 21.5,

HCO3 = 248, SO4 = 221.8, Cl = 79.2

Find the total hardness, carbonate hardness and non-carbonate hardness.

Solution:

Hardness due to M++ as CaCo3 = hardness as M++ (mg/lit) x eq. wt. of CaCo3 / eq. wt. of M++

Total hardness = [ 65x 50/20 +51 x 50/12.2] mg/lit as CaCo3.

= 371.52 mg/lit as CaCo3

Alkalinity due to B as CaCo3 = alkalinity as B– x eq. wt. of CaCo3 / eq. wt. of B

= 248 x 50/61

= 203.28 mg/lit

In this case.

If, TH > A

then, CH = A

NCH = TH-CH

= 371.52-203.28

= 168 mg/lit

The analysis of a sample of water shows the following results in mg/lit:

 Ca Mg Na K No3 Cl HCo3 7 12 20 30 12 40 68

The concentration of Sr (Stroncium) is eq. to the hardness of 2.52 mg/lit as CaCo3 and the carbonate alkalinity in this water is zero. Calculate the total hardness and NCH in mg/lit as CaCo3.

Solution,

Total hardness = Ca++ x 50./20 +Mg++ x 50/12.2 +2.52

= 7×50/20 + 12×50/12.2 +2.52

= 69.2 mg/lit as CaCo3.

Bicarbonate alkalinity= 68 x 50/61 =55.74 mg/lit as CaCo3

In this case,

TH>A

CH= A

NCH = TH-CH

=69.2-55.74

= 13.46 mg/lit as CaCo3

3. The total hardness value obtained from the complete analysis of a water sample is found to be 150 mg/lit. The analysis further showed that the concentrations of all the three principal cations causing hardness are numerically the same. If the value of carbonate hardness is 77 mg /lit, calculate;

1) the value of NCH

2) the concentrations of principal cations; and

3) the value of total alkalinity in mg/lit.

Solution:

Total Hardness = Carbonate hardness + Non Carbonate Hardness

NCH = 150 – 77 = 73 mg/lit

Let, P is the concentration of principle cations.

Now, TH= P x 50/20 + P x 50/12.2 + P x 50/43.8

Principle concentrations (P)= 19.38 mg/lit

In this case, NCH is greater than zero so total hardness is greater than
alkalinity.

When, TH > A

then, CH = A = 77 mg/lit as CaCo3

NCH = TH-CH

 Read Also: Quantity of Water Numerical

4. The hardness of a water sample was found to be 300 mg/lit as CaCo3. The hardness was found due to Ca and Mg ions being equal in Water. The analysis showed the concentration of HCO3, was 150 mg/lit,

ii) the concentration of Ca and Mg

ii) the value of alkalinity of water.

iii) the CH and NCH; and

Solution:

i. Let, P is the concentration of principal cations.

Hardness due to M++ as CaCo3 = hardness as M++ (mg/lit) x eq. wt. of CaCo3 / eq. wt. of M++

Total Hardness = [50P/20 + 50P/12.2] mg/lit as CaCo3

300 = 6.6 P

Therefore, P = 45.45 mg/lit

ii) Bicarbonate alkalinity

Bicarbonate alkalinity = 150 x 50/61 = 122.95 mg/lit as CaCO3

Total alkalinity = CA + BA = 0 + 122.95 = 122.95 mg/lit

iii) Here, Total hardness > Total alkalinity

Carbonate hardness = Total alkalinity = 122.95 mg/lit

Total Hardness = Carbonate hardness + Non Carbonate Hardness NCH

= 300-122.95

= 177.05 mg/lit as CaCO

5. If 400 ml of water with a pH 6 is mixed with 700 ml of water with a pH of 8, what will be the resultant pH of the mixture?

Given,

Va = 400 ml, (PH)a = 6

Vh = 700 ml, (pH)b = 8

Total volume of solution = 1100 ml

[H+] = antilog10 [-pH] ,

pH = log10 1/[H+]

[Hla = 10-6 moles/lit,

Conc. of [H+]a in 400 ml = 10-6 x 400 / 1100 = 3.64×10-7 moles/lit

Conc. of [H+lb in 700 ml = 10-8 x 700/1100 = 6.36×10-9 moles/lit

Concentration mix
= 3.64 x 10-7 + 6.36 x 10-9

= 3.7036 x 10-7 moles/lit

pH= 6.43

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