Answer:
2.28 × 10^-3 mol/L
Explanation:
The equation for the equilibrium is
CN^- + H2O ⇌ HCN + OH^-
Ka = 4.9 × 10^-10
KaKb = Kw
4.9 × 10^-10 Kb = 1.00 × 10^-14
Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5
Now, we can set up an ICE table
CN^- + H2O ⇌ HCN + OH^-
I/(mol/L) 0.255 0 0
C/(mol/L) -x +x +x
E/(mol/L) 0.255 - x x x
Ka = x^2/(0.255 - x) = 2.05 × 10^-5
Check for negligibility
0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255
x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6
x = sqrt(5.20 × 10^-6) = 2.28 × 10^-3
[OH^-] = x mol/L = 2.28 × 10^-3 mol/L
- The substance that made up of a particular mixture can be regarded as Solute and solvent
- A mixture can be regarded as material which is made up of more than one different Chemical substance or substances , whereby they are not chemically combined.
- Substance could be any matter such as <em>salt, rice, and or compounds</em>
- A mixture can be regarded as physical combination involving two or more substances whereby each entities identities are retained, then mixed in the form such as ;
- <em>solutions</em>
- <em> suspensions </em>
- <em> colloids</em>.
- A mixture could be in this form <em>(Solute + solvent = solution)</em>
Therefore, the substance that made up of a particular mixture can be regarded as <em>Solute and solvent</em>
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