ahhh im confused!!
Calculate the pH of a solution formed by mixing 372 mL of a solution containing 4.4 x 10-6 M NaOH with 286 mL of 6.0 x 10-2 M NaOH. Report your answer to 2 decimal places.
Are both of these NaOH? If so, calculate mols of 1, mols of the other, add mols, and divide mols/L(total volume) = molarity.
Then pOH = -log (OH^-)
and pH + pOH = 14.
To calculate the pH of the solution, we need to consider the concentration of hydroxide ions (OH-) in the solution. We can find the combined concentration of hydroxide ions by adding the individual concentrations of the two solutions.
First, let's determine the moles of hydroxide ions in each solution:
For the first solution, we have:
Volume = 372 mL = 372/1000 = 0.372 L
Molarity (M) = 4.4 x 10^-6 M
Moles of NaOH = Molarity x Volume = 4.4 x 10^-6 Moles/L x 0.372 L
For the second solution, we have:
Volume = 286 mL = 286/1000 = 0.286 L
Molarity (M) = 6.0 x 10^-2 M
Moles of NaOH = Molarity x Volume = 6.0 x 10^-2 Moles/L x 0.286 L
Now, we can calculate the total moles of hydroxide ions by adding the moles from both solutions.
Total moles of hydroxide ions = Moles of NaOH (first solution) + Moles of NaOH (second solution)
Next, we can calculate the total volume of the combined solution by adding the volumes of both solutions.
Total volume of the combined solution = Volume (first solution) + Volume (second solution)
Finally, we can calculate the concentration of hydroxide ions in the combined solution by dividing the total moles by the total volume.
Concentration of hydroxide ions = Total moles of hydroxide ions / Total volume of the combined solution
Now, to find the pOH of the solution:
pOH = -log[OH-]
And finally, to calculate the pH, we can use the following equation:
pH = 14 - pOH
Plug in the previously calculated pOH value into the equation to get the pH of the solution.
Please follow the steps above to calculate the pH of the given solution.