Hey there!
Given the reaction:
B + H⁺ => HB⁺
At half-equivalence point : [B] = [HB⁺]
=> [B] / [HB⁺] = 1
Henderson-Hasselbalch equation :
pH = pKa + log ( [B] ) / ( HB⁺)]
pH = 14 - pKb + log ( 1 )
pH = 14 - 7.95 + 0
pH = 6.05
Answer C
Hope that helps!
The pH at the half-equivalence point in the titration of a weak base with a strong acid is 6.05.
The given parameters;
pKb = 7.95
The half-equivalence point reaction is given as;
[tex]HB \ ---> B^{-} \ + \ H^+[/tex]
Apply Henderson-Hasselbalch equation as shown below;
[tex]pH = pK_b \ + \ log_{10} (\frac{B^-}{HB} )\\\\[/tex]
At half-equivalence point: [tex]B^- = HB[/tex]
[tex]pH = 14 \ - pK_b + log(1)\\\\pH = 14 -7.95 + 0\\\\pH = 6.05[/tex]
Thus, the pH at the half-equivalence point in the titration of a weak base with a strong acid is 6.05.
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