<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq)
+ Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>)
= 0.021 M.
Ka(HCN) = 4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷
4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻]
/ [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] =
x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M
- x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>
pH solution = 8.89
<h3>Further explanation</h3>
Given
The concentration of HBr solution = 1.3 x 10⁻⁹ M
Required
the pH
Solution
HBr = strong acid
General formula for strong acid :
[H⁺]= a . M
a = amount of H⁺
M = molarity of solution
HBr⇒H⁺ + Br⁻⇒ amount of H⁺ = 1 so a=1
Input the value :
[H⁺] = 1 x 1.3 x 10⁻⁹
[H⁺] = 1.3 x 10⁻⁹
pH = - log [H⁺]
pH = 9 - log 1.3
pH = 8.89
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Symbols on the periodic table represent an element.
Cl is also known as chlorine.
Answer : The heat of reaction for the process is, 1374.7 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The main chemical reaction is,

The intermediate balanced chemical reaction will be,
(1)

(2)

(3)

We reversing reaction 1, 3 and multiplying reaction 2 by 2 and then adding all the equations, we get :
(1)

(2)

(3)

The expression for heat of reaction for the process is:



Therefore, the heat of reaction for the process is, 1374.7 kJ