Answer:
Determine the pH of the solution half-way to the end-point on the pH titration curve for acetic acid.
Explanation:
The equation for the ionization of acetic acid is
HA + H₂O ⇌ H₃O⁺ + A⁻
For points between the starting and equivalence points, the pH is given by the Henderson-Hasselbalch equation:
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D)
At the half-way point, half of the HA has been converted to A⁻, so [HA] = [A⁻]. Then,

The pKₐ is the pH at the half-way point in the titration.
Answer:
Explanation:
First we look generally at what makes K2SO4.
In one mole of K2SO4, there are 2 moles of the potassium ion (K+) and 1 mole of sulfate ion (SO4 2-).
Knowing that; in 1.75 moles of K2SO4, there must be 2 x 1.75moles of potassium ion (K+) and 1 x 1.75moles of Sulfate ion (SO4 2-)
This gives us 3.5moles of K+ and 1.75moles of SO4 2-
Iron. Well actually steel, but iron is the closest.
<span>B) N2(g) + 3H2(g) → 2NH3(g)
</span>I hope this helps :)
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