Determine the pH of a solution containing 0.045 M NaOH and 0.025 M KI:
With activities:
Without activities:
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To determine the pH of a solution containing NaOH and KI, we need to consider the dissociation of their ions in water and the subsequent effect on the pH.
With activities:
1. Write the balanced chemical equations for the dissociation of NaOH and KI:
NaOH → Na+ + OH-
KI → K+ + I-
2. Since NaOH is a strong base, it completely dissociates into Na+ and OH- ions. However, KI is a strong electrolyte but does not affect the pH significantly as it doesn't produce OH- or H+ ions.
3. As a result, the OH- ions from NaOH will contribute to the alkalinity of the solution. The concentration of OH- ions can be calculated from the concentration of NaOH.
4. Calculate the concentration of OH- ions:
[OH-] = [NaOH] = 0.045 M
5. Since OH- is a hydroxide ion and its concentration affects the pH, we can calculate the pOH using the formula pOH = -log[OH-].
6. Calculate the pOH:
pOH = -log(0.045) ≈ 1.35
7. To determine the pH, we can use the fact that pH + pOH = 14. Rearrange the equation to calculate pH:
pH ≈ 14 - pOH ≈ 14 - 1.35 ≈ 12.65
Therefore, the pH of the solution containing 0.045 M NaOH and 0.025 M KI is approximately 12.65.
Without activities:
Since KI does not react with water to produce H+ or OH- ions, its presence does not significantly affect the pH of the solution. Therefore, the pH of the solution will mainly be determined by the concentration of OH- ions from NaOH.
To calculate the pH:
1. Determine the concentration of OH- ions based on the concentration of NaOH: [OH-] = [NaOH] = 0.045 M.
2. Calculate the pOH using the formula pOH = -log[OH-]: pOH ≈ -log(0.045) ≈ 1.35.
3. Use the fact that pH + pOH = 14 to determine the pH: pH ≈ 14 - pOH ≈ 14 - 1.35 ≈ 12.65.
Therefore, the pH of the solution containing 0.045 M NaOH and 0.025 M KI is approximately 12.65.