Answer:
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.
<em>Moles of NaOH
</em>
Moles of NaOH = 25 × 0.173
Moles of NaOH = 4.32 mmol
===============
<em>Moles of HCl
</em>
Moles of HCl = 35 × 0.342
Moles of HCl = 12.0 mmol
===============
<em>Amount of excess reactant
</em>
NaOH + HCl ⟶ NaCl +H₂O
<em>n</em>/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
===============
<em>Concentration of the excess HCl
</em>
Total volume = 25 + 35
Total volume = 60 mL
<em>c </em>= millimoles HCl/millilitres HCl
<em>c</em> = 7.6/60
<em>c</em> = 0.13 mol/L
===============
<em>Calculate the pH
</em>
The HCl dissociates completely to hydronium ions, so
[H₃O⁺] = 0.13 mol·L⁻¹
pH = -log[H₃O⁺]
pH = -log0.13
pH = 0.89
