Question

NaOH (s) was added to 1.0L of HClO4 (aq) 0.49 M.

a) Calculate [H3O+] in the solution after the addition on 0.11 mol of NaOH (s)

b) Calculate [H3O+] in the solution after the addition of 0.84 mol of NaOH (s)

Answer 1

Solution:

The reaction between NaOH and HClO4 is given as:

NaOH + HClO4 = NaClO4 + H2O

From above equation, it can be seen that 1 mol of NaOH is required to neutralize 1 mol of HClO4.

The number of moles of HClO4 in 1L of 0.49 M solution = Molarity x Volume in L = 0.49 M x 1L = 0.49 mol

Part A)

When 0.11 mol of NaOH is added, then it is neutralized by 0.11 mol of HClO4, hence number of moles of HClO4 left in solution = 0. 49 mol - 0.11mol = 0.38 mol

Hence, molarity = 0.38 mol / 1.0 L = 0.38 M

[H+] = [H3O+] = 0.38 M

Part B)

When 0.84 mol of  NaOH is added, then 0.49 mol of it is neutralize by 0.49 mol of HClO4.

Therefore, number of mol of NaOH left = 0.84 - 0.49 =0.35 mol

Hence, [OH-] = 0.35 M and then pOH = -log [OH-]

pOH = - log 0.35 = -log 35 x 10^-2 = 2 - log 35

= 2 - 1.544 = 0.456

Since, pH + pOH = 14

Hence, pH = 14 - pOH = 14 - 0.456 = 13.544

[H+] = [H3O+] = 10^-pH = 10^-13.544

[H3O+] = 2.88 x 10^-14 M