Question

1.A student performs a titration by despensing 24.6 ml of 0.10 M NaOH into a 35.5 mL sample of HCL. Calculate the molarity, mol/L, of the HCL solution.

2.Write the balanced equation for the dissolution of potassium hydrogen phthalate, KHC8H4O4, also known as KHP, in water to produce potassium ion and hydrogen phthalate ion, HC8H4O4-.

3. Write overall balanced equation for the reaction between sodium hydroxide, NaOH, and hydrogen phthalate ion, HC8H4O4-.

4.Write the net reaction that represents acid-base neutralization.

Answer 1

1) The balanced chemical equation is

NaOH + HCl ------> NaCl + H₂O

As per the stoichiometric reaction,

1 mole NaOH = 1 mole HCl.

Moles of NaOH = (24.6 mL)*(1 L/1000 mL)*(0.10 mol/L) = 0.00246 mole.

Therefore, moles of HCl neutralized = 0.00246 mole.

Molar concentration of HCl = (0.00246 mole)/[(35.5 mL)*(1 L/1000 mL)] = 0.06929 mol/L ≈ 0.0693 mol/L (ans).

2) The dissolution equation is given by

KHC₈H₄O₄ (aq) --------> K⁺ (aq) + HC₈H₄O₄^- (aq)

3) The reaction between NaOH and HC₈H₄O₄^- is a 1:1 molar equation and the balanced chemical equation is

NaOH (aq) + HC₈H₄O₄^- (aq) ------> NaC₈H₄O₄^- (aq) + H₂O (l)