testing bronsted lowry reaction predictions...... im super stuck can someone help me steo by step how to figure this out assuming i use the 5 step process?
write a balanced chemical equation for ch3coona and hcl reaction
position of equilibrium.
I know that ch3coo is a SB
and Hcl is a SA
so reactants are favored and shown as <50% but i cant fiigure out how to write the equation.
Have you been told that ch3coona isn't anything in chemistry. You need to find the caps key for CO, Co and co mean something different just as m and M aren't the same.
I don't have your table of acidic/base values but you do this. You will need to fill in the numbers and decide where th equilibrium is.
CH3COONa ==> CH3COO^- + Na^+
CH3COO^- is a strong base. Na^+ is a spectator ion. CH3COO^- accepts a proton to form CH3COOH.
HCl ==> H3O^+ + Cl^-
H3O^+ is a strong acid. Cl^- is a spectator ion. H3O^+ donates a proton.
CH3COO^- + H3O^+ ==> CH3COOH
..SB........SA.........WA
This reaction is far to the right.
no i have to teach myself and that's why i was so confused i knew CH3COO would gain a proton but i didn't realize that hcl changed to HCl ==> H3O^+ + Cl^- so it makes a little more sense.I do wanna make sure i understand and i have troubles breaking down the entities. i may ask for more help but only to make sure i am on the right track.
Thank you sooooooo much for helping though
I suppose to make it complete the HCl reacts with H2O as follows:
HCl + H2O ==> H3O^+ + Cl^-
You're quite welcome and come back anytime you have a question. It helps us help you if you explain in detail what you don't understand.
To write a balanced chemical equation for the reaction between CH3COONa and HCl, you can follow these steps:
Step 1: Identify the reactants and products.
In this case, the reactants are CH3COONa (sodium acetate) and HCl (hydrochloric acid). The products of the reaction will be sodium chloride (NaCl) and acetic acid (CH3COOH).
Step 2: Write the unbalanced equation.
Based on the reactants and products identified, the unbalanced equation can be written as follows:
CH3COONa + HCl → NaCl + CH3COOH
Step 3: Balance the equation.
Now, you need to balance the equation by making sure that the number of atoms of each element is the same on both sides of the equation. Start by balancing the sodium (Na) atoms. Since there's one Na on the left side and one on the right side, it is already balanced. Next, balance the hydrogen (H) atoms by adding a coefficient of 2 in front of HCl:
CH3COONa + 2HCl → NaCl + CH3COOH
Finally, balance the carbon (C) and oxygen (O) atoms. Since there is only one carbon atom and two acetate ions (CH3COO-) on the left side, you'll need to add a coefficient of 2 in front of CH3COOH to balance the carbon atoms. This will also balance the oxygen atoms:
CH3COONa + 2HCl → NaCl + 2CH3COOH
The balanced equation for the reaction between CH3COONa and HCl is:
CH3COONa + 2HCl → NaCl + 2CH3COOH
Step 4: Determine the position of equilibrium.
To determine the position of equilibrium, you would need additional information such as the concentrations or amounts of the reactants and the conditions (temperature, pressure, etc.) under which the reaction is occurring.
Step 5: Analyze the equilibrium.
Using the provided information that CH3COO- is a weak base (WB) and HCl is a strong acid (SA), we can infer that the reaction will favor the formation of the products (NaCl and CH3COOH) since weak bases and strong acids react to form their respective salt and weak acid. Therefore, the position of equilibrium would be shifted towards the products in this reaction.
I hope these steps help you with understanding how to write a balanced chemical equation and determine the position of equilibrium. If you have any further questions, feel free to ask!