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Calculate the alkalinity for
each water sample.  Show your
calculations.  Summarize the data in a figure

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Ground water

alkalinity (mg/L as CaCo3)= A  x N x50000




=32.5ml N =0.02mg/L , sample(ml)=100ml

converting into liters



alkalinity(mg/L  as CaCO3)=  0.0325*0.02*50000

———————-   =325mg/L



Unknown water sample

   Here A=1.2ml covert to liter =0.0012L

N =
0.02mg/l sample ml=100ml =0.1L

alkalinity (mg/L as CaCo3)=(0.0012*0.02*50000)/0.1=12mg/L


Surface water


A=14.5mL   N=0.02mg/L, sample(mL)=100mL=0.1L

                    By Using




What is one source of error for
this analysis and how can it be avoided?

ANS. Sources of error may as follows:

                        1 Spills of water.

          2 Uncleaned beakers

 3 Visual end point dependably
somewhat proportionality point.

4 Mixing of water samples by using same beaker .

5 stirer may cause error .


Based on your knowledge of the
chemistry of different sources of water, explain the differences in alkalinity
that you found.  Where the results what
you expected? Comment on the unknown water sample.  Where might this water source have come from?


ANS High alkalinity of ground water is
because of the way that it has irrelevant creation of carbonic of carbonic
acids or other acidic materials. while then again surface water has continually
confronting expansion of rain water,pesticides,organics leading to its less


You are working in a lab that
routinely does alkalinity analyses.  You
usually buy 0.02 N sulfuric acid from Hach, but this time your boss orders pure
sulfuric acid from Sigma-Aldrich.  You
have to make 5 L of 0.02 N sulfuric acid to do a series of alkalinity
titrations.  Describe how you would make
the solution.  Show all calculations.



 we need equivalent mass of
sulphuric acid which is 49g

volume = 5L

desired normality = 0.02N

Grams of compound needed = (N desired)(Equivalent mass)(vol. in Lt.

(0.02)(49)(5) = 4.9g


A0.02N solution required 4.9g of a pure sulphuric acid powder
diluted to 5L. But the acid is a liquid and it is not one hundred percent pure
active sulphuric acid. we will need to calculate what volume of the concentrate
acid that contains 4.9g of sulphuric acid.

Formula :  vol. of acid needed
= grams of acid needed/percent conc.*speific gravity



percent conc. of sulphuric acid =97%

vol. needed=(4.9)/(0.97*1.84)=2.74m

so, if we take 2.74ml of conc. sulphuric acid and diluted to 5L, we
would have 0.02N sulphuric acid solution.



     5.Explain how alkalinity
is different from hardness.


Ans Alkalinity measures of capacity of an
answer for kill corrosive without changing ph to control and keep up water ph
hardness estimated in degrees ppm of calcium carbonate.


6.What is the difference between
phenophtalein alkalinity and total alkalinity?

Ans   A phenolpthalein marker is added to the
sample.a bromcresol green methyl red pointer is added and titration continous
to the second end point at a ph between 4.3 to 4.9 this esteem demonstrates add
up to alkalinity and is a deliberate of all carbonate,bicarbonate and hydroxide
substance in water. Phenolpthalein changes shading.

7.Why is alkalinity important for drinking
water treatment?  What might happen to
the treatment processes if the alkalinity of the water was low?  How would this impact treatment?

Alkalinity is imperative on the grounds that without alkalinity water is more
probable be inclined to be acidic alkalinity stablize water ph. Low alkalinity
is very serious and xcan cause rapid fluctuation in ph water can not easily
counteract effects of acid its not easily to balance ph . its very difficult to
treat water which is low in alkalinity.

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