If 22 mL of 0.50 M HCl is added to 112 mL of 0.20 M NaOH, what is the final pH?

Question

Quizplus · Accepted Answer

First we need to find the moles of HCl and NaOH:

moles of HCl = (0.50 mol/L) x (0.022 L) = 0.011 mol
moles of NaOH = (0.20 mol/L) x (0.112 L) = 0.022 mol

Since NaOH is a strong base and HCl is a strong acid, they will completely react with each other in a 1:1 ratio:

HCl + NaOH → NaCl + H2O

The limiting reagent in this reaction is HCl, which will completely react with 0.011 mol of NaOH. This will leave 0.011 moles of excess NaOH in solution.

To find the concentration of the excess NaOH, we need to subtract the moles of NaOH that reacted with HCl from the total moles of NaOH:

moles of excess NaOH = 0.022 mol - 0.011 mol = 0.011 mol

To find the concentration of the excess NaOH, we need to divide the moles by the total volume of the solution:

[OH-] = moles of excess NaOH / total volume
[OH-] = 0.011 mol / 0.134 L
[OH-] = 0.082 mol/L

Now we can find the pOH of the solution using the equation:

pOH = -log[OH-]
pOH = -log(0.082)
pOH = 1.08

Finally, we can find the pH of the solution using the equation:

pH + pOH = 14
pH = 14 - pOH
pH = 14 - 1.08
pH = 12.92

Therefore, the final pH is C) 12.93.

If 22 ML of 0