Calculate the pH of the resulting solution if 23.0 mL of 0.230 M HCl(aq) is added to 28.0 mL of 0.230 M NaOH(aq) .

pH = 12.35

Explanation:

The HCl reacts with NaOH, thus:

HCl + NaOH → NaCl + H₂O

1 mole of HCl reacts per mole of NaOH

To know the pH we must know the moles of NaOH and the moles of HCl that reacts

Moles NaOH and HCl

NaOH: 0.0280L * (0.230mol / L) = 6.44x10⁻³ moles

HCl: 0.0230L * (0.230mol / L) = 5.29x10⁻³

As moles NaOH > Moles HCl, the moles that remains of NaOH are:

Moles NaOH = 6.44x10⁻³ - 5.29x10⁻³ = 1.15x10⁻³ moles NaOH = Moles OH⁻

In 23.0mL + 28.0mL = 51.0mL = 0.0510L:

1.15x10⁻³ moles OH⁻ / 0.0510L =

0.0225M OH⁻

pOH = -log [OH⁻]

pOH = 1.647

And pH = 14 - pOH

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