Question

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Answer 1

1.

NaOH volume = 24.32 ml = 0.02432 L

NaOH concentration = 0.134 M = 0.134 mol/L

$\therefore$ no. of moles of NaOH = 0.02432 L × 0.134 mol/L = 3.2589 mmol

2.

HCl + NaOH $\rightarrow$ NaCl + H₂O

Stoichiometric balance shows HCl : NaOH = 1:1

So, 3.2589 mmol reacts with 3.2589 mmol of HCl to reach endpoint.

3.

Volume of HCl × molarity of HCl = moles of HCl

molarity of HCl = 3.2589 mmol / 10 ml = 0.326 M