What is the pH of a solution formed by the combination of 700 mL of 0.20 M HCl with 300 mL of 0.50 M NaOH?Question 5Select one:a.0.699b.1.70c.2.00d.0.854e.12.0

First, we need to determine the moles of HCl and NaOH.

Moles of HCl = Molarity * Volume = 0.20 M * 0.700 L = 0.14 moles Moles of NaOH = Molarity * Volume = 0.50 M * 0.300 L = 0.15 moles

NaOH is a strong base and HCl is a strong acid. When they react, they will neutralize each other according to the equation: HCl + NaOH -> NaCl + H2O

Since we have more moles of NaOH than HCl, the solution will be basic after the reaction. The excess moles of NaOH = 0.15 moles - 0.14 moles = 0.01 moles.

The total volume of the solution is 700 mL + 300 mL = 1.0 L.

The concentration of NaOH in the solution is therefore 0.01 moles / 1.0 L = 0.01 M.

The pH of a solution is given by the formula pH = -log[H+]. However, since we have a basic solution, we need to find the pOH first and then subtract from 14 to find the pH (since pH + pOH = 14 at 25 degrees Celsius).

The pOH is given by -log[OH-]. Since NaOH is a strong base, [OH-] is equal to the concentration of NaOH, which is 0.01. Therefore, pOH = -log(0.01) = 2.

Finally, subtracting the pOH from 14 gives the pH of the solution: pH = 14 - pOH = 14 - 2 = 12.

So, the correct answer is e. 12.0.