Answer to Question #162094 in General Chemistry for Shraddha

Question #162094

Q.3100 mL of a water sample required 25mL of N/50H_(2)SO_(4) for neutralization to phenolphthalein end point.After this methyl orange indicator was added and further acid required was again 25mL Calculate type and extent of alkalinity.


1
Expert's answer
2021-02-09T04:40:46-0500

100 ml of water up to phenolphthalein end point = (25ml of N ) / (50 H2SO4)

100 ( Np) = 25 ( N/50 )


Np = "\\frac{25}{100}" ("\\frac{N}{50}" )


strength of alkalinity up to phenolphthalein end point in terms of CaCO3

= Np (50)(1000) ppm


P = "\\frac{25}{100}" ("\\frac{1}{50}" ) 50 (1000)pm = 250 ppm


now 100ml of water to methyl orange end point

= 25 + 3 = 28 ml of N/(50H2SO4)

100(Nm) = 28 ("\\frac{N}{50}" )

Nm = "\\frac{28}{100}" ("\\frac{1}{50}")


stength ofalkalinity up to mrthyl orange end point in terms of CaCO3 equivalent hardness

= Nm (50)(1000)

= "\\frac{28}{100}" ("\\frac{1}{50}" ) 50 (1000) = 280 ppm


As P>(1/2)M , hence OH- and C0"_3^{2-}" are present

Alkalinity due to OH- = 2P - M = 2(250) - 280 = 220 ppm

Alkalionity due to C0"_3^{2-}" = 2(M-P) = 2(280 - 250) = 60 ppm

total Alkalinity = 220 + 60 = 280 ppm



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