Answer:
The concentration of NaOH solution is 3,65M
Explanation:
The neutralization of NaOH with HCl is:
NaOH + HCl → NaCl + H₂O
You are titrating 12,67 mL HCl 2,74M with your NaOH solution, the moles of HCl in the beaker are:
12,67 mL ≡ 0,01267L×[tex]\frac{2,74 moles}{1L}[/tex] = 0,0347 moles HCl ≡ 0,0347 moles NaOH
The volume that you require for total neutralization of 12,67 mL HCl 2,74M is:
13,65 mL - 4,50 mL = 9,50 mL ≡ 0,0095L
Thus, the concentration of NaOH solution is:
[tex]\frac{0,0347 moles}{0,0095L}[/tex] = 3,65M
I hope it helps!