B I hope it’s right I don’t really help a lot but yeah lol
<span>The reason it will be 7 for some titrations is that when you titrates a strong acid with a strong base for example HCl and NaOH the salt formed is conjugate base of strong acid and will be a very weak base
That means that it cannot produce any OH^-1 and all the H+ has been converted to water.The only source of H+ or OH is water with a Ka of 10^-14 so the pH = -log [H+]=-log 10^-7 = 7
second reason is
When you titrates a weak acid with strong base at equivalence point
only a water solution of the conjugate base exists
CH3COOH + NaOH ----- Na+ CH3COO^-1 + H2O
Since the conjugate base is the conjugate base of a weak acid it will hydrolyze in water like so
for instance Na+ CH3COO^-1 + HCl---- CH3COOH + NaCl the equivalence point will be way BELOW 7 and in the case of above will be less than 5. So pH of 7 at equivalence point is only reached in strong acid strong base titrations.
hope this helps</span>
Answer: :))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
Explanation:
This reaction is most likely to fall under SN2 because the
thing called carbonication does not occur in SN1. The carbon forms a partial
bond with the nucleophile during the intermediate phase and the leaving group.
So for this question the reaction will fall under SN2.
