254.88 mL of an NaOH solution with an analytical concentration of 0.143 M is goi
ID: 908512 • Letter: 2
Question
254.88 mL of an NaOH solution with an analytical concentration of 0.143 M is going to be titrated with an HCl solution with an analytical concentration of 0.0924 M.
A. What is the pH of a resulting solution when the above NaOH solution is titrated with 115.66 mL of the above HCl solution?
B. What is the pH of a resulting solution after the NaOH solution has been titrated with a total of 941.06 mL of the HCl solution?
C. How many mL of the HCl solution must be added to the original NaOH solution in order to exactly reach the equivalence point?
D. What is the pH of the solution at the equivalence point?
Explanation / Answer
The balanced chemical equation for the given reaction was
NaOH + HCl => NaCl + H2O
That means the two reactants will react in 1:1 ration to form 1 mole of NaCl
No. of moles of NaOH taken = 0.143 moles/L x 0.25488 L = 0.03644 moles
A) If 115.66 ml of acid was added
No. of moles of acid = 0.0924 mole/L x 0.11566 L = 0.01068 moles
No. of moles of NaOH will left after reaction = 0.03644 - 0.01068 = 0.02576
Final NaOH concentration = 0.02576 / 0.37054 = 0.06952 M = [OH-]
pOH = -log[OH-] = 1.1578
Hence pH = 14-pOH = 12.8421
A) If 941.06 ml of acid was added
No. of moles of acid = 0.0924 x 0.94106 = 0.08695
No. of moles of acid will left after reaction = 0.08695 - 0.03644 = 0.05051
Concentration of HCl = 0.05051 / 1.19594 = 0.04223 M = [H+]
pH = -log[0.04223] = 1.3743
C) To reach the equivalence point, no. of moles of acid required = no. of moles of base = 0.03644
Volume of HCl= 0.03644 moles / 0.0924 moles.L-1 = 394.37 ml
D) At equivalence point there will not be any H+ or OH- ions hence pH = 7 (Neutral)
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