Solution Found!
We will abbreviate malonic acid, CH2(CO2H)2, as H2M. Find the pH and concentrations of
Chapter 9, Problem 9-5(choose chapter or problem)
We will abbreviate malonic acid, \(\mathrm{CH}_{2}\left(\mathrm{CO}_{2} \mathrm{H}\right)_{2}\), as \(\mathrm{H}_{2} \mathrm{M}\). Find the \(\mathrm{pH}\) and concentrations of \(\mathrm{H}_{2} \mathrm{M}, \mathrm{HM}^{-}\), and \(\mathrm{M}^{2-}\) in
(a) \(0.100 \ \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{M}\);
(b) \(0.100 \ \mathrm{M} \ \mathrm{NaHM}\);
(c) \(0.100 \ \mathrm{M} \ \mathrm{Na}_{2} \mathrm{M}\).
Questions & Answers
QUESTION:
We will abbreviate malonic acid, \(\mathrm{CH}_{2}\left(\mathrm{CO}_{2} \mathrm{H}\right)_{2}\), as \(\mathrm{H}_{2} \mathrm{M}\). Find the \(\mathrm{pH}\) and concentrations of \(\mathrm{H}_{2} \mathrm{M}, \mathrm{HM}^{-}\), and \(\mathrm{M}^{2-}\) in
(a) \(0.100 \ \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{M}\);
(b) \(0.100 \ \mathrm{M} \ \mathrm{NaHM}\);
(c) \(0.100 \ \mathrm{M} \ \mathrm{Na}_{2} \mathrm{M}\).
ANSWER:Step 1 of 3
(a)
The equilibrium reaction of the malonic acid is as follows.
For this reason, assume that a solution of 
Let’s write the equilibrium expression for the given reaction as follows.
Assume that,
And 
Substitute these values in expression (3)
Let’s calculate the concentration of hydronium ion.
Let’s calculate the pH of the solution.
Let’s write the equilibrium reaction for the equation (2).
Let’s rearrange the equation (4)
Substitute the given values.
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