Question

3. Calculate in the titrant. 4. Another 1,00 mL of the NaOH solution is added after the equivalence point. Calculate e reaction mixture. in the 5. Suppose 10.00 mL of water was added to the HCl solution before the titration was begun. How would your answers to questions 1-3 change?

Answer 1

(1)

The number of moles of H₃O⁺ = 0.2376mol/L × (5.00×10^-3 L)

= 1.188×10^-3 mol. (Answer)

(2)

H₃O⁺ + OH^-$\rightarrow$ 2 H₂O

The number of moles of OH^- = number of moles of H₃O⁺ = 1.188×10^-3 mol. (Answer)

(3)

[OH^-] = [NaOH] = The number of moles / volume of solution

= 1.188×10^-3mol/(40.02×10^-3 L) = 0.02969 mol/L

c_OH^- = 2.97×10^-2 M. (Answer)

(4)

1.00ml of NaOH is added beyond the equivalent point .

The moles of OH^- in excess taken = 1.00×10^-3L × 2.97×10^-2mol/L = 2.97×10^-5mol

The volume of solution mixture = 5.00ml + 40.02ml + 1.00ml= 46.02 ml

[OH^-] = 2.97×10^-5/(46.02×10^-3L)

= 6.45×10^-4 M. (Answer)