<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>
Answer:
Suppose you added some solid NaCl to a saturated solution of NaCl at 20℃ and warmed the mixture to 40℃. What would happen to the added NaCl?
Explanation:
can you help with this one
Answer: A pattern of same atomic orbitals can be seen about elements in the same period with respect to electron structures.
Explanation:
The horizontal rows in a period table are called periods.
Elements present in the same period will have same atomic orbitals.
For example, electronic distribution of Na is 2, 8, 1 and it is a third period element.
Similarly, electronic distribution of Cl is 2, 8, 7 and it is also a third period element.
Hence, both Na and Cl will have K, L, M shells, that is, they have three atomic orbitals.
Thus, we can conclude that a pattern of same atomic orbitals can be seen about elements in the same period with respect to electron structures.
