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The ph at the equivalence point of 25.00ml of 0.3m hf titrated with 0.3m NaOH where ka value us 6.6 x 10^ -4 is calculated to be....??? since moles are equal, f ion will react with oh ion how do you calculate ph at equivalence
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At the equivalence point, all the weak acid (HF) has reacted with the strong base (NaOH) to form its conjugate base (F-). The pH is determined by the hydrolysis of the F- ion. Here's how to calculate it:
1. Calculate the concentration of F- at equivalence:
- Moles of HF initially: 0.025 L
0.3 mol/L = 0.0075 moles
- Moles of NaOH at equivalence = moles of HF = 0.0075 moles
- Volume at equivalence: 25.00 mL + 25.00 mL = 50.00 mL = 0.050 L
- [F-] = 0.0075 moles / 0.050 L = 0.15 M
2. Set up the ICE table for the hydrolysis of F-:
F- + H2O ⇌ HF + OH-
I 0.15 0 0
C -x +x +x
E 0.15-x x x
3. Write the Kb expression for the hydrolysis:
Kb = [HF][OH-] / [F-]
Since Kb
Ka = Kw (1.0 x 10^-14)
Kb = 1.0 x 10^-14 / 6.6 x 10^-4 = 1.5 x 10^-11
4. Solve for x:
1.5 x 10^-11 = (x)(x) / (0.15 - x)
Since Kb is very small, we can approximate (0.15 - x) ≈ 0.15
x^2 = 2.25 x 10^-12
x = 1.5 x 10^-6
5. Calculate the pOH:
pOH = -log(1.5 x 10^-6) = 5.82
6. Calculate the pH:
pH = 14 - pOH = 14 - 5.82 = 8.18
Therefore, the pH at the equivalence point is approximately 8.18.
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