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A 1. 30 m solution of the weak acid ha is created. if the ka of the weak acid is 3. 06×10−9, what is the ph of the solution? the equilibrium expression is:____.

User IronWaffleMan
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1 Answer

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Final answer:

The pH of a 1.30 M solution of a weak acid HA with a Ka of 3.06×10⁻⁹ is found using an ICE table and the acid dissociation equilibrium expression. After determining the concentration of hydronium ions, we calculate the pH by taking the negative logarithm of the hydronium ion concentration.

Step-by-step explanation:

To determine the pH of a 1.30 M solution of the weak acid HA with a Ka value of 3.06×10⁻⁹, we use an ICE table (Initial, Change, Equilibrium) to process the acid dissociation equilibrium:

HA(aq) + H₂O(l) ⇒ H3O+(aq) + A-(aq)

Assuming an initial concentration of 1.30 M for HA and 0 M for H3O+ and A-, we denote the change in concentration due to dissociation as 'x'. At equilibrium, the concentrations are (1.30 - x) M for HA, 'x' M for H3O+ and A-. As weak acids dissociate minimally, we typically assume x to be very small relative to the initial concentration, so 1.30 - x is approximately 1.30 M.

The equilibrium expression is:

Ka = [H3O+][A-]/[HA]

For the weak acid HA, this equation becomes:

3.06×10⁻⁹ = x² / (1.30)

By solving for 'x' and then using the relationship pH = -log[H3O+], we can find the pH of the solution.

User Smbat Poghosyan
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