Answer:
Explanation:
<u>According to Arrhenius concept of acid and base:</u>
"When a base in a solution, produces/yields OH- (Hydroxide) ions."
So, when a base is dissolved in a solution, it produces OH- ions.
<u>For example:</u>
NaOH ⇄ Na⁺ + OH⁻ (So, it is a base)
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
Wouldn't be a then not c if both were solutions.
I want to say 8 but i'm not sure. I just did a lesson on this and my notes say it could be 8 but then again I'm not sure.
Hello!
When NaOH is added to a buffer composed of CH₃COOH and CH₃COO⁻, the following reactions happen:
-First, the NaOH is neutralized by CH₃COOH:
NaOH + CH₃COOH → H₂O + CH₃COONa
-Second, the CH₃COONa dissociates in its ions:
CH₃COONa → CH₃COO⁻ + Na⁺
-Finally, CH₃COO⁻ (a weak base) reacts with water to form OH⁻ ions and regenerate CH₃COOH
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻
By this sequence of reactions, the buffer can mitigate the effect of the strong base added.
Have a nice day!
Answer:
-1.78 V
Explanation:
There are several rules required to calculate the cell potential:
- given standard cell potential, we may reverse the equation: the products of a given reaction become our reactants, while reactants become our products in the reversed equation. For a reversed equation, we change the sign of the cell potential to the opposite sign;
- if we multiply the whole equation by some number, this doesn't influence the cell potential value. It only produces a different expression in the equilibrium constant.
That said, notice that the initial reaction with respect to the final reaction is:
- reversed: chromium(III) cation and chloride anion become our reactants as opposed to the products in the initial reaction, so we change the sign of the cell potential to a negative value of -1.78 V;
- each coefficient is multiplied by a fraction of
. It doesn't influence the value of the cell potential.
Thus, we have a cell of E = -1.78 V.
