Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻)
= 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) =
5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.
<u>Answer:</u> The pressure that must be applied to the apparatus is 0.239 atm
<u>Explanation:</u>
To calculate the osmotic pressure, we use the equation for osmotic pressure, which is:
or,
where,
= osmotic pressure of the solution
i = Van't hoff factor = 1 (for non-electrolytes)
= mass of sucrose = 3.40 g
= molar mass of sucrose = 342.3 g/mol
= Volume of solution = 1 L
R = Gas constant = 
T = temperature of the solution = ![20^oC=[20+273]K=293K](https://tex.z-dn.net/?f=20%5EoC%3D%5B20%2B273%5DK%3D293K)
Putting values in above equation, we get:
Hence, the pressure that must be applied to the apparatus is 0.239 atm
Opp that that person has to correct answer
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