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.
The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
<h3>Determine the freezing point of the solution </h3>
First step : Calculate the molality of NaCl
molality = ( 2.5 grams / 58.44 g/mol ) / ( 230 * 10⁻³ kg/ml )
= 0.186 mol/kg
Next step : Calculate freezing point depression temperature
T = 2 * 0.186 * kf
where : kf = 1.86°c.kg/mole
Hence; T = 2 * 0.186 * 1.86 = 0.69°C
Freezing point of the solution
Freezing temperature of solvent - freezing point depression temperature
0°C - 0.69°C = - 0.69°C
Hence the Freezing temperature of the solution is - 0.69°C
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<u>Answer:</u>
<em>The molarity of the 
solution is 
</em>
<em></em>
<u>Explanation:</u>
The Balanced chemical equation is
Mole ratio of : KCl is 1 : 1
So moles = moles KCl
So Molarity
= 0.000402M or mol/L is the Answer
(Or) is the Answer
It's a base because it increases the concentration of hydroxid ions
