Final answer:
To determine the molarity of a sodium acetate solution with a pH of 9.70, we calculate the pOH from the given pH and then find the hydroxide ion concentration. Because sodium acetate dissociates completely, this concentration also represents the molarity of the sodium acetate, which is 10⁻¹³·³⁰ M.
Step-by-step explanation:
To find the molarity of a sodium acetate solution with a pH of 9.70, we can use the formula for the dissociation of acetic acid (CH3COOH), since sodium acetate (NaCH3COO) is its conjugate base.
The dissociation of acetic acid in water can be represented as:
CH3COOH(aq) ⇌ CH3COO⁻(aq) + H⁺(aq)
Sodium acetate in water will hydrolyze to form acetic acid and hydroxide ions (OH⁻).
The pH is related to the pOH by the equation pH = 14 - pOH, and the pOH is related to the hydroxide ion concentration ([OH⁻]) by the equation pOH = -log[OH⁻]. Hence, we first need to calculate the pOH:
pH = 9.70
pOH = 14.00 - 9.70 = 4.30
Now we calculate the concentration of OH⁻ ions:
[OH⁻] = 10⁻¹³·³⁰ M
Since sodium acetate completely dissociates in water, the concentration of acetate ions will be equal to the concentration of hydroxide ions. Therefore, the molarity of the sodium acetate solution is also going to be 10⁻¹³·³⁰ M.