Question

In the titration of 10.00 ml of sodium hydroxide with 0.1500 M hydrochloric acid, an equivalence volume of 15. 25 mL is obtained using a pH meter to monitor the titration.

(a) If the same titration were carried out using an acid-base indicator with an acid dissociation constant of 1.58 x 10^-5 , what would be the titrant volume required to get to the endpoint

Answer 1

we know that at endpoint mole of NaOH is equal to moles of HCl

So, M1V1 = M2V2 ( where M1 and M2 = molarity of NaOH & HCl, V1 & V2 = volume of NaOH & HCl)

from the first titration, we can calculate the molarity of NaOH

M1 * 10 = 0.15 * 15.25

M1 = 0.228 M

In second titration we have given dissociation constant (K) = 1.58 * 10^-5

HCl --------------> H⁺ + Cl^-

at time 't' 0.15 - x x x

therefore K = [ H⁺ ] [ Cl^-] / [HCl]

1.58 * 10^-5 = x * x / 0.15 - x

x ² = 1.58 * 10^-5 * 0.15 ( dissociation constant is very low so we neglect the x in denominator term)

x = 1.53 * 10 ^-3 M

therefore at end point

M1V1 = M2V2

0.228 * 10 = 1.53 * 10 ^-3 * V2

V2 = 1.4 * 10 ^-3 ml = 1.4 litre ------------- answer

1.4litre volume of HCl required to get end point.

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