Acid Dissociation Equilibrium Problem: Consider the dissociation equilibrium of

Question

Acid Dissociation Equilibrium Problem:

Consider the dissociation equilibrium of an acid in water,

For acetic acid at room temperature, the dissociation equilibrium pK value is 4.76, i.e., means cA?cH+ /cA = 10?4.76 mol/l. For the dissociation products of water, cOH?cH+ = 10?14 mol2/L2 . If 1 mol of the acid A is dissolved in 10 L water, find the pH value of the solution, and the concentration cA? of the dissolved acid. (Hint: Use the constraint from overall charge neutrality in the solution.)

Acid Dissociation Equilibrium Problem: Consider the dissociation equilibrium of an acid in water, A OH + H . For acetic acid at room temperature, the dissociation equilibrium pK value is 4.76, i.e., means cACH+ /CA = 10^4.76 mol/l. For the dissociation products of water, COH CH+ = 10^14 mol^2/L^2 . If 1 mol of the acid A is dissolved in 10 L water, find the pH value of the solution, and the concentration CA of the dissolved acid. (Hint: Use the constraint from overall charge neutrality in the solution.) A + H , H2O

Explanation / Answer

Kwater= 10-14 = [H3O+][OH-]

initial   [H+] = 1*10-7 M because water is neutral.

Now, you're adding acid. Let's find how much:

added [H+] = 1 mol / 10 L = 0,1 M. There's no reaction here, just dissociation, so you're just increasing the amount of H+. Add this new value to the initial one and you'll notice that the initial is almost zero regarding to the added one.

The new pH is:

pH = -log ([H+]) = -log(0.1) = 1

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