Respuesta :
Answer:
i) The initial pH before adding any HCl solution is 13.06
ii) The volume of acid required to reach equivalence point is 28.75 mL
iii) The pH after addition of 5.0 mL HCl is 12.899
iv) The pH at the equivalence point is 7
Explanation:
i) Here we have the concentration of the sample = 0.115M NaOH
Therefore pH of NaOH
pOH = -log[OH⁻] = -log[0.115] = 0.9393
pH + pOH = 14
∴ pH = 14 - pOH = 14 - 0.9393 = 13.06
The initial pH before adding any HCl solution = 13.06
ii) The equivalence point is the point of complete neutralization, therefore;
NaOH + HCl → NaCl + H₂O
One mole of NaOH reacts with one mole of HCl to form one mole of NaCl and one mole of H₂O
Therefore, 0.115 Mole of NaOH will be completely neutralized by 0.115 mole of HCl, hence we have
25 × 0.115 moles of NaOH ≡ x × 0.1 of HCl
Where:
x = Volume of HCl required in mL
∴ x = 25×0.115/0.1 = 28.75 mL of HCl is required to reach equivalence point
The volume of acid required to reach equivalence point = 28.75 mL
iii) The pH after addition of 5.0 mL HCl is found as follows
The volume, y of NaOH that is completely neutralized by 5.0 mL HCl is given by the following expression
y×0.115 = 5×0.1
y = 5×0.1/0.115 = 4.35 mL
Therefore, since the total volume is now 25 + 5 = 30 mL and the amount of 0.115 M NaOH in the solution is 25 - 4.35 = 20.65 mL, the molarity of the solution is therefore;
20.65 mL/1000 × 0.115 = 0.002375 moles in 30 mL
Number of moles in one liter = 0.002375/(30/1000) = 0.0792 moles
Hence the new molarity of NaOH = 0.0792 M
The pH = 14 - pOH = 14 - (-log[OH])
= 14 - (-log(0.0792)) = 12.899
The pH after addition of 5.0 mL HCl = 12.899
iv) The pH at equivalence point after complete neutralization is the pH of a neutral solution hence
pH at the equivalence point = 7.
