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.12.0Clear my choice

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 NaOH is in excess (0.15 moles > 0.14 moles), we will have some leftover NaOH after the reaction.

Moles of leftover NaOH = Moles of NaOH - Moles of HCl = 0.15 moles - 0.14 moles = 0.01 moles

The total volume of the solution is the sum of the volumes of HCl and NaOH = 0.700 L + 0.300 L = 1.00 L

The molarity of the leftover NaOH = Moles / Volume = 0.01 moles / 1.00 L = 0.01 M

The pH of a solution of a strong base is determined by the pOH, which is calculated as pOH = -log[OH-].

Since NaOH is a strong base, it will dissociate completely in water, so the concentration of OH- ions is equal to the molarity of NaOH = 0.01 M.

So, pOH = -log(0.01) = 2

The pH is then calculated as pH = 14 - pOH = 14 - 2 = 12

So, the pH of the solution is 12 (option e).