Calculate the pH of 250 ml of water after mixing with 250 ml of 5mM NaOH.
5 mM x (250/500) = 2.5 mM = 0.0025 M
Then (H^+)(OH^-)= 1E-14 and solve for H^+; convert to pH. In pure H2O the (OH^-) is 1E-7 but that isn't enough to significantly affect the total OH^-.
To calculate the pH of the solution after mixing water with NaOH, we need to consider the dissociation of NaOH in water. NaOH dissociates into Na+ and OH- ions.
First, let's calculate the number of moles of NaOH that were added to the solution:
Moles of NaOH = concentration of NaOH x volume of NaOH
Moles of NaOH = 5 mM x 0.25 L
Moles of NaOH = 0.00125 moles
Since NaOH dissociates into equal amounts of Na+ and OH- ions, we have 0.00125 moles of OH- ions in a total volume of 0.5 L (250 ml + 250 ml).
Now, let's calculate the concentration of OH- ions in the solution:
Concentration of OH- ions = moles of OH- ions / volume of solution
Concentration of OH- ions = 0.00125 moles / 0.5 L
Concentration of OH- ions = 0.0025 M
Since the OH- ions are responsible for the basic nature of the solution, we can use the pOH scale to calculate the pOH of the solution:
pOH = -log10[OH-]
pOH = -log10(0.0025)
pOH = 2.60
Finally, to calculate the pH of the solution, we can use the equation:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 2.60
pH = 11.40
Thus, the pH of the solution after mixing water with 5mM NaOH is approximately 11.40.
To calculate the pH of the resulting solution after mixing water with NaOH, we need to consider the dissociation of NaOH in water.
Firstly, we need to determine the concentration of NaOH in the final solution. Since 250 ml of 5mM NaOH is added to 250 ml of water, the total volume of the solution is 500 ml (250 ml + 250 ml).
To find the final concentration of NaOH, we use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of NaOH (5mM)
V1 = initial volume of NaOH (250 ml)
C2 = final concentration of NaOH
V2 = final volume of the solution (500 ml)
Plugging in the values:
(5 mM) * (250 ml) = C2 * (500 ml)
C2 = (5 mM * 250 ml) / 500 ml
C2 = 2.5 mM
Now that we have the final concentration of NaOH, we can proceed to calculate the pH. NaOH is a strong base that fully dissociates into hydroxide ions (OH-) and sodium ions (Na+).
The balanced equation for the dissociation of NaOH is:
NaOH → Na+ + OH-
Since 1 mole of NaOH dissociates into 1 mole of hydroxide ions, the concentration of hydroxide ions in the solution will be 2.5 mM.
The pH can be calculated using the equation:
pOH = -log[OH-]
pOH = -log[2.5 mM]
Using the logarithm base 10 (common logarithm), we can find the negative logarithm of 2.5 mM to determine the pOH.
pOH = -log(2.5)
Next, we need to convert pOH to pH using the equation:
pH = 14 - pOH
pH = 14 - (-log(2.5))
Finally, we can calculate the pH of the solution by substituting the values into the equation:
pH = 14 - (-log(2.5))
With the values substituted and calculations performed, we find the pH of the solution after mixing 250 ml of water with 250 ml of 5mM NaOH.