What is the ph of the solution formed when 25 ml of 0.173 m naoh is added to 35 ml of 0.342 m hcl?

0.89

Explanation:

You are mixing an acid and a base, so there will be a neutralization reaction.

One of the reactants will be in excess, so we must determine its concentration and then calculate the pH.

Moles of NaOH

Moles of NaOH = 25 × 0.173

Moles of NaOH = 4.32 mmol

Moles of HCl

Moles of HCl = 35 × 0.342

Moles of HCl = 12.0 mmol

Amount of excess reactant

              NaOH + HCl ⟶ NaCl +H₂O

n/mmol:   4.32      12.0

The 4.32 mmol of NaOH reacts completely with 4.32 mmol of HCl.

Excess HCl = 12.0 – 4.32

Excess HCl = 7.6 mmol

Concentration of the excess HCl

Total volume = 25 + 35

Total volume = 60 mL

c = millimoles HCl/millilitres HCl

c = 7.6/60

c = 0.13 mol/L

Calculate the pH

The HCl dissociates completely to hydronium ions, so

[H₃O⁺] = 0.13 mol·L⁻¹

pH = -log[H₃O⁺]

pH = -log0.13

pH = 0.89

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