Leo carefully pipets 50.0 mL of 0.500 M NaOH into a test tube. She places the test tube into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.250 M HCl into
another test tube. When Leo reaches to put this test tube of acid into the beaker along
with test tube of base she accidentally knocks the test tubes together hard enough to
break them and their respective contents combine in the bottom of the beaker. Is the
solution formed from the contents of the two test tubes acidic or basic? What is the pH of
the resulting solution?
Please answer below questions one by one to assist you receive full credits
(Alternatively, you can discard my hints below, solve the problem using your own way
and send me the picture/copy of your complete work through email)
The mole of NaOH before mixing is
mol (save 3 significant figures)
The mole of HCl before mixing is
mol (save 4 significant figures)
After mixing, the solution is
(choose from acidic or basic)
The total volume of mixture is
L (save 3 significant figures)
The concentration of [OH-] is
M (save 3 significant figures)
The concentration of [H'l is
M (save 3 significant figures)

Let's consider the neutralization reaction between HCl and NaOH.

NaOH + HCl ⇒ NaCl + H₂O

To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.

NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol

HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol

Now, let's determine the reactant in excess and the remaining moles of that reactant.

                    NaOH    +     HCl ⇒ NaCl + H₂O

Initial           0.0200       0.0188

Reaction    -0.0188       -0.0188

Final         1.20 × 10⁻³          0

The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³  moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:

[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M

NaOH is a strong base according to the following equation.

NaOH ⇒ Na⁺ + OH⁻

The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.

The pOH is:

pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02

We will calculate the pH using the following expression.

pH = 14.00 - pOH = 14.00 - 2.02 = 11.98

The pH is 11.98. Since pH > 7, the solution is basic.

You can learn more about neutralization here:

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the answer is barium and copper bromide