Hello!
First, we need to determine the concentration of KHP. This concentration is:
[tex]\frac{3,5g KHP }{50 mL sol}* \frac{1000 mL}{1 L}* \frac{1 mol KHP}{204,22 g KHP}=0,343 M [/tex]
The neutralization reaction is:
KHP + NaOH → NaKP + H₂O
At the equivalence point:
[tex]moles NaOH=moles KHP \\ \\ cNaOH*vNaOH=cKHP*vKHP \\ \\ cNaOH= \frac{cKHP*vKHP}{vNAOH}= \frac{0,343M*50mL}{(24,65 mL-1,85 mL)} =0,75 M [/tex]
So, the concentration of NaOH is 0,75M
Have a nice day!