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: brainly.com/question/16255996
