Attached problem Question 7You have a solution of a diprotic acid H2A (Kaj=3.10x103, Ka2=2.37x10-6) with aformal concentration of 0.250 M. Please calculate the concentrations of OH", HA",H2A and A2 in this solution.

Given, Molar concentration of H2A = 0.250 M First dissociation constant of H2A = Ka1 = 3.10 x 10-3…

Answered: Question 7 You have a solution of a…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Step 1

Given,

Molar concentration of H₂A = 0.250 M

First dissociation constant of H₂A = K_a1 = 3.10 x 10^-3

Second dissociation constant of H₂A = K_a2 = 2.37 x 10^-6

[OH^-] = ?

[HA^-] = ?

[H₂A] = ?

[A²^-] = ?

Step 2

ICE table of H₂A is :

..................................H₂A (aq) <---------------------------> H⁺ (aq).....................+......................HA^- (aq)

Initial (I)..................0.250 M...............................................................0.0 M................................................0.0 M

Change (C)................-0.250α.............................................................+0.250α.......................................+0.250α

Equilibrium (E)..........0.250(1-α) M.....................................................0.250α M......................................0.250α M

Here,

α = Amount dissociated per mole

Expression of K_a1 is :

$K_{a 1} = \frac{[H^{+}] . [H A^{-}]}{[H_{2} A]} S u b s t i t u t e t h e v a l u e s : 3.10 x 10^{- 3} = \frac{{(0.250 α)}^{2}}{0.250 (1 - α)} 3.10 x 10^{- 3} = \frac{{0.250 α}^{2}}{(1 - α)} 0.250 α^{2} + 3.10 x 10^{- 3} α - 3.10 x 10^{- 3} = 0$

On solving the quadratic equation :

α = 0.1053

So,

[H⁺] = [HA^-] = 0.250α M

= 0.250 x 0.1053 M

= 0.0263 M

Also,

[H₂A] = 0.250(1-α) M

= 0.250 (1-0.1053) M

= 0.224 M

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