Respuesta :
Answer is: pH of solution of sodium cyanide is (c) 11.30.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻)
= 0.25 M.
pKa(HCN) = 9.21.
Ka(HCN) = 10∧(-9.21).
Ka(HCN) = 6.16·10⁻¹⁰.
Ka(HCN) · Kb(CN⁻) = 10⁻¹⁴.
Kb(CN⁻) = 10⁻¹⁴ ÷
6.16·10⁻¹⁰
Kb(CN⁻) = 1.62·10⁻⁵.
Kb = [HCN] · [OH⁻]
/ [CN⁻].
[HCN] · [OH⁻] =
x.
[CN⁻] = 0.25 M - x..
1.62·10⁻⁵ = x² / (0.25 M
- x).
Solve quadratic equation: x = [OH⁻] = 0.002 M.
pOH = -log(0.002 M) = 2.70.
pH = 14 - 2.70 = 11.30.
11.30 is the pH of a 0.25 m solution of sodium cyanide and this value shows that the solution is basic.
How we calculate the pH of the given solution?
We can calculate the pH by using the formula pH = 14 - pOH.
For this we will calculate the value of pOH as pOH = -log[OH]
And value of concentration of OH will be calculated by using the formula of Kb for the below reaction:
Cyanide ion + water ⇄ Hydrogen cyanide + hydroxide ion
In the question given that:
Value of pKa = 9.21
Ka = 10⁻⁹°²¹
Ka = 6.17 × 10⁻¹⁰
And Kb = Kw/Ka
Kw for water = 1.0 × 10⁻¹⁴, then
Kb = 1.0 × 10⁻¹⁴ / 6.17 × 10⁻¹⁰ = 1.621 × 10⁻⁵
For the above reaction, Kb will be written as:
Kb = [Hydroxide ion] [Hydrogen cyanide] / [Cyanide ion]
According to the ICE table:
[Hydroxide ion] = [Hydrogen cyanide] = x
[Cyanide ion] = 0.25 - x or equivalent to 0.25
On putting these values on the above equation we get,
Kb = x² / 0.25
1.621 × 10⁻⁵ = x² / 0.25
x = 2.013 × 10⁻³
or [OH⁻] = 2.013 × 10⁻³
Now we calculate pOH by putting this value as:
pOH = -log[2.013 × 10⁻³] = 2.70
And pH is calculated as:
pH = 14 - 2.70 = 11.30
Hence, option (c) is correct i.e. 11.30 is the pH of the sodium cyanide solution.
To know more about pH, visit the below link:
https://brainly.com/question/13557815
