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In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 2.85 by using the green arrows adjacent to the pH value indicated on the probe in the solution. Once you adjust the pH, note the corresponding OH? ion concentration in M as given in the graphic on the left side of the simulation. Make sure to select the option "Concentration (mol/L)" above the graphic. Select on the Logarithmic scale below the graphic.Find the pOH of the solution.Express your answer numerically to two decimal places.

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Answer:

pOH = 11.15

[OH⁻] = 7.079 × 10⁻¹² M

Explanation:

The pH of the solution is 2.85. We can find out pOH using the following relation:

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 2.85 = 11.15

From the definition of pOH we can look for [OH⁻].

pOH = -log [OH⁻]

- pOH = log [OH⁻]

[OH⁻] = antilog (- pOH) = antilog (-11.15) = 7.079 × 10⁻¹² M

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