Hi, here is a basic summary of what we did in a lab; there were 3 reactions: The procedure: Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. NaOH(s)-> Na+(aq) + OH-(aq) ΔH1=-34.121kJ Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of HCl to form water and an aqueous solution of sodium chloride. NaOH(s) + H+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH2=-83.602kJ Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of HCl to form water an an aqueous solution of sodium chloride. H+(aq) + OH-(aq) + Na+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH3= -50.2kJ The ΔH values were calculated by dividing the heat gained by the number of moles (each reaction had 0.05moles of NaOH) The problem: Net ionic equations for reaction 2 & 3: 2: NaOH(s) + H+(aq) -> H2O + Na+(aq) 3: H+(aq) + OH-(aq) -> H2O i) In reaction 1, ΔH1 represents the heat evolved as solid NaOH dissolves. Look at the net ionic equations for reactions 2 and 3 and make similar statements as to what ΔH2 and ΔH3 represent. ii) Compare ΔH2 with (ΔH1 + ΔH3). Explain in sentences the similarity between these two values by using your answer to #5 above. Attempt at answering: i) Firstly, ΔH2 represents the heat evolved as the hydrogen ion displaces the sodium ion, creating a single displacement reaction. ΔH3 represents the heat evolved as the hydrogen and hydroxide ion form water via a neutralization reaction. ii) ΔH2 is equal to (or supposed to be, this is a source of error while calculating) (ΔH1 + ΔH3). The similarity between these two values is that .. (this is where I get confused!)
Source https://www.physicsforums.com/threads/calorimetry-help-chemistry.399653/
Answer:
B and C
Explanation:
When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:
![pKa_1=-Log[6.9x10^-^3]=2.16](https://tex.z-dn.net/?f=pKa_1%3D-Log%5B6.9x10%5E-%5E3%5D%3D2.16)
![pKa_2=-Log[6.2x10^-^8]=7.20](https://tex.z-dn.net/?f=pKa_2%3D-Log%5B6.2x10%5E-%5E8%5D%3D7.20)
![pKa_3=-Log[4.8x10^-^13]=12.31](https://tex.z-dn.net/?f=pKa_3%3D-Log%5B4.8x10%5E-%5E13%5D%3D12.31)
The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which 
is the <u>acid</u> and 
is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.
Answer:
Step 3
Explanation:
I am back sorry it took me so long, I believe its Step 3 because that's were you chose your strategy, you use it in Step 4, but if its not Step 3 its Step 4
1. identifying
problem solving step: 1)__ the problem is most difficult: not in habit of asking what the problem really is; in the habit of reacting or giving up
2. representing
problem solving step: 2) __ the problem: abstract or external representation
3. strategy
problem solving step: 3) selecting an appropriate __; trial and error vs means-ends analysis
4. implementing
problem solving step: 4) __ the strategy; dependent on previous steps
