# Hardness and Alkalinity Numerical | Water Supply Engineering

Table of Contents

In this article, we will discuss Hardness and Alkalinity Numerical.

** 1. Formula **

Alkalinity due to B^{–} as CaCo_{3} = alkalinity as B^{– }x eq. wt. of CaCo_{3 }/ eq. wt. of B^{–} |

Hardness due to M^{++} as CaCo_{3} = hardness as M^{++} (mg/lit) x eq. wt. of CaCo_{3 }/ eq. wt. of M^{++} |

Total Hardness = Carbonate hardness + Non carbonate hardness |

[H^{+}] = antilog_{10} [-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,

HCO_{3} = 248, SO_{4} = 221.8, Cl = 79.2

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

Solution:

**Hardness due to M ^{++} as CaCo_{3} = hardness as M^{++} (mg/lit) x eq. wt. of CaCo_{3 }/ eq. wt. of M^{++}**

Total hardness = [ 65x 50/20 +51 x 50/12.2] mg/lit as CaCo_{3}.

= 371.52 mg/lit as CaCo_{3}

**Alkalinity due to B ^{–} as CaCo_{3} = alkalinity as B^{– }x eq. wt. of CaCo_{3 }/ 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 | No_{3} | Cl | HCo_{3} |

7 | 12 | 20 | 30 | 12 | 40 | 68 |

The concentration of Sr (Stroncium) is eq. to the hardness of 2.52 mg/lit as CaCo_{3} and the carbonate alkalinity in this water is zero. Calculate the total hardness and NCH in mg/lit as CaCo_{3}.

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 CaCo_{3}.

Bicarbonate alkalinity= 68 x 50/61 =55.74 mg/lit as CaCo_{3}

In this case,

TH>A

CH= A

NCH = TH-CH

=69.2-55.74

= 13.46 mg/lit as CaCo_{3}

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 CaCo_{3}

NCH = TH-CH

Read Also: Quantity of Water Numerical |

4. The hardness of a water sample was found to be 300 mg/lit as CaCo_{3}. 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 CaCo_{3} = hardness as M^{++} (mg/lit) x eq. wt. of CaCo_{3 }/ eq. wt. of M^{++}**

Total Hardness = [50P/20 + 50P/12.2] mg/lit as CaCo_{3}

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

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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^{+}] = antilog_{10} [-pH] ,

pH = log_{10} 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**

Civil Engineer & CEO of Naba Buddha Group