Calculate the pH of 175.0 mL of H2O following the addition of 0.57 mL of 1.39 M NaOH
To calculate the pH of a solution, we first need to determine the concentration of the acid or base in the solution. In this case, we have added 0.57 mL of 1.39 M NaOH to 175.0 mL of H2O.
To calculate the concentration of NaOH in the final solution, we can use the following formula:
C1V1 = C2V2
Where:
C1 = initial concentration of NaOH
V1 = initial volume of NaOH
C2 = final concentration of NaOH
V2 = final volume of the solution
Plugging in the given values:
(1.39 M)(0.57 mL) = C2(175.0 mL + 0.57 mL)
Simplifying the equation:
0.793 = C2(175.57)
Dividing both sides by 175.57:
C2 ≈ 0.004513 M
Now we have the concentration of NaOH in the solution. Since NaOH is a strong base, it dissociates completely in water, producing hydroxide ions (OH-).
To find the concentration of hydroxide ions (OH-) in the solution, we can use the formula for molarity:
Molarity (M) = moles of solute / volume of solution (in liters)
Since NaOH dissociates into one hydroxide ion (OH-) per molecule, the concentration of hydroxide ions will be equal to the concentration of NaOH.
Concentration of hydroxide ions (OH-) = 0.004513 M
To calculate the pOH of the solution, we can use the formula:
pOH = -log10(OH- concentration)
pOH = -log10(0.004513)
pOH ≈ 2.35
Finally, to find the pH of the solution, we can use the formula:
pH + pOH = 14
pH + 2.35 = 14
pH ≈ 11.65
Therefore, the pH of the solution after the addition of 0.57 mL of 1.39 M NaOH to 175.0 mL of H2O is approximately 11.65.