What is the pH and pOH a solution that was made by adding 400 mL of water to 350 mL of 5.0 x 10^-3 M NaOH solution?
This is just a dilution problem.
5E-3M NaOH x [(350)/(350+400)] OR
use the dilution formula of
mL1 x M1 = mL2 x M2
To find the pH and pOH of the solution, we first need to calculate the concentration of the resulting solution.
The initial concentration of NaOH is given as 5.0 x 10^-3 M. Since 400 mL of water is added, the final volume of the solution becomes 750 mL (350 mL NaOH + 400 mL water).
To calculate the final concentration, we'll use the formula:
M1V1 = M2V2
where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
Plugging in the values:
(5.0 x 10^-3 M)(350 mL) = M2(750 mL)
Solving for M2:
M2 = (5.0 x 10^-3 M)(350 mL) / 750 mL
≈ 0.00233 M
Now that we have the concentration of the solution, we can find the pOH and pH.
The pOH is calculated using the formula:
pOH = -log[OH-]
Since NaOH is a strong base and dissociates completely, the concentration of OH- will be equal to the concentration of NaOH, which is 0.00233 M.
pOH = -log(0.00233)
≈ 2.633
To find the pH, we can use the fact that in a neutral solution, the pH and pOH add up to 14:
pH + pOH = 14
pH = 14 - pOH
= 14 - 2.633
≈ 11.367
Therefore, the pH of the solution is approximately 11.367 and the pOH is approximately 2.633.