Answer: 10
Step-by-step explanation:
Since 10.0 cm of a 1.00 x 10^-2 mol dm-3 aqueous solution of sodium hydroxide is added to a volumetric flask and the total volume is made up to 1.00 dm³ with distilled water, we have effectively diluted the solution.
To determine the concentration of sodium hydroxide in the diluted solution, we can use the dilution formula:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, C1 = 1.00 x 10^-2 mol dm-3, V1 = 10.0 cm = 10.0 x 10^-3 dm³ (converted to dm³), C2 is what we need to find, and V2 = 1.00 dm³.
Plugging in the values:
(1.00 x 10^-2 mol dm-3)(10.0 x 10^-3 dm³) = C2(1.00 dm³)
C2 = (1.00 x 10^-2 mol dm-3)(10.0 x 10^-3 dm³) / (1.00 dm³)
C2 = 1.00 x 10^-4 mol dm-3
Now that we have the concentration of sodium hydroxide in the diluted solution, we can determine the concentration of hydroxide ions (OH-) using stoichiometry. Since sodium hydroxide dissociates in a 1:1 ratio, the concentration of hydroxide ions is also 1.00 x 10^-4 mol dm-3.
To calculate the pOH of the solution, we can use the formula:
pOH = -log10 [OH-]
pOH = -log10 (1.00 x 10^-4)
pOH = 4
Since pH + pOH = 14 (at 25°C), we can find the pH of the solution by subtracting the pOH from 14:
pH = 14 - 4
pH = 10
Therefore, the pH of the diluted solution is 10.