How many moles of NaOH must be added to 1.00L of this solution to increase the PH to 9.25?
You need to read your question. You haven't posted all of it.
To determine how many moles of NaOH must be added to the solution, we need to use the concept of pH and the dissociation of NaOH in water.
Step 1: Determine the concentration of OH- ions needed to achieve a pH of 9.25.
Since the solution is in water, we can use the equation: pH + pOH = 14.
pOH = 14 - pH = 14 - 9.25 = 4.75
Step 2: Convert pOH to OH- concentration.
pOH is defined as the negative logarithm (base 10) of the molar concentration of OH- ions.
We can use the formula: pOH = -log[OH-], where [OH-] is the concentration of OH- ions.
To find [OH-], we need to take the antilog of pOH.
[OH-] = 10^(-pOH) = 10^(-4.75)
Step 3: Determine the volume of the solution.
The given information states that the volume is 1.00 L.
Step 4: Calculate the moles of NaOH needed.
Since NaOH is a strong base, it completely dissociates in water.
The stoichiometry between NaOH and OH- is 1:1, meaning one mole of NaOH produces one mole of OH- ions.
To calculate the moles of NaOH, we can multiply the volume of the solution (in liters) by the concentration of OH- ions (in moles per liter).
Moles of NaOH = volume of solution (in liters) * [OH-] (in moles per liter)
Putting it all together:
Moles of NaOH = 1.00 L * 10^(-4.75)