.012 mol of solid NaOH is added to a 1 L solution of HCl 0.01 M. What is the pH of the solution?

mols HCl = M x L = 0.01 x 1 = 0.01

mols NaOH = 0.12
mols NaOH after neutralization = 0.11.
M NaOH = M OH^- = mols/L.
Then pOH = -log(OH^-) and
pH + pOH = pKw = 14.
Given pOH and pKw, solve for pH.

To find the pH of the solution, we need to determine the concentration of the resulting H+ ions after the reaction between NaOH and HCl takes place.

First, let's calculate the number of moles of HCl in the 1 L solution:
Moles of HCl = Concentration of HCl × Volume of Solution
Moles of HCl = 0.01 M × 1 L
Moles of HCl = 0.01 mol

Since the reaction between NaOH and HCl is a 1:1 ratio, the number of moles of HCl consumed will be equal to the number of moles of NaOH added.

Now, let's calculate the concentration of the remaining HCl in the solution after the reaction:
Remaining moles of HCl = Initial moles of HCl - Moles of NaOH
Remaining moles of HCl = 0.01 mol - 0.012 mol
Remaining moles of HCl = -0.002 mol

Since we have a negative value for the remaining moles of HCl, it means that all the HCl has been consumed in the reaction. Therefore, there are no HCl molecules left in the solution to dissociate and produce H+ ions.

As a result, the pH of the solution after adding 0.012 mol of solid NaOH to a 1 L solution of HCl 0.01 M would be determined by the dissociation of the NaOH. NaOH is a strong base that fully dissociates in water, releasing OH- ions.

Since the concentration of remaining HCl is zero, we can calculate the concentration of OH- ions from the NaOH:
Concentration of OH- ions = Moles of NaOH / Volume of Solution
Concentration of OH- ions = 0.012 mol / 1 L
Concentration of OH- ions = 0.012 M

To find the pOH of the solution, we can use the formula:
pOH = -log[OH-]
pOH = -log(0.012)
pOH ≈ 1.92

Finally, to find the pH of the solution, we can use the relation:
pH + pOH = 14
pH = 14 - pOH
pH ≈ 14 - 1.92
pH ≈ 12.08

Therefore, the pH of the solution after adding 0.012 mol of solid NaOH to a 1 L solution of HCl 0.01 M is approximately 12.08.