30.0 mL of 1.50 M NaOH is diluted to 50.0 mL. What is the resulting pH?
mL x M = mL x M
Solve for M NaOH.
That equals (OH^-).
pOH = -log(OH^-)
Then pH + pOH = pKw = 14
Solve for pH.
To find the resulting pH, we need to calculate the concentration of NaOH after dilution and then use that concentration to determine the hydroxide ion concentration ([OH-]).
To calculate the concentration after dilution, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case:
C1 = 1.50 M (initial concentration)
V1 = 30.0 mL (initial volume)
C2 = ? (final concentration)
V2 = 50.0 mL (final volume)
Plugging in the values:
1.50 M * 30.0 mL = C2 * 50.0 mL
Calculating C2:
C2 = (1.50 M * 30.0 mL) / 50.0 mL
C2 = 0.90 M
After dilution, the concentration of NaOH is 0.90 M.
Now, we can determine the hydroxide ion concentration ([OH-]) from the concentration of NaOH (0.90 M) since NaOH is a strong base that fully dissociates in water:
[OH-] = concentration of NaOH
Therefore, [OH-] = 0.90 M
To calculate the pOH, we use the formula:
pOH = -log [OH-]
pOH = -log (0.90 M)
Using a calculator, we find:
pOH = 0.046
Finally, to find the pH, we use the equation:
pH = 14 - pOH
pH = 14 - 0.046
Calculating the pH:
pH = 13.954
The resulting pH after dilution is approximately 13.954.