Answer to Question #211312 in General Chemistry for Hina

Question

A water sample is alkaline to both phenolphthalein and methyl orange. 0.50 mlof this sample requires 15 ml of H2SO4 for phenolphthalein indicator and another 10 ml for complete neutralization. Calculate the type of alkalinity in ppm.

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