Answer to Question #162094 in General Chemistry for Shraddha

Question

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.

Answer 1

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

100 ( N_p) = 25 ( N/50 )

N_p = $\frac{25}{100}$ ($\frac{N}{50}$ )

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

= N_p (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(N_m) = 28 ($\frac{N}{50}$ )

N_m = $\frac{28}{100}$ ($\frac{1}{50}$)

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

= N_m (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