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)
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
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