could some one please help with this titration calculation? (1 Viewer)

[fayad158]

Dilute 10ml of vinegar to 250ml. then four 25ml aliquots of the diluted vinegar were titrated against 0.1M NaOH (average Volume needed is 23ml). find the concentration of the original vinegar.

any help would be apreciated, i think i can do it but i didn' t use civi=cfvf, i times the n of NaOH by 250/25.

thanks

 

[fayad158]

anyone?

 

[brenton1987]

Dilute 10ml of vinegar to 250ml. then four 25ml aliquots of the diluted vinegar were titrated against 0.1M NaOH (average Volume needed is 23ml). find the concentration of the original vinegar.

any help would be apreciated, i think i can do it but i didn' t use civi=cfvf, i times the n of NaOH by 250/25.

CH₃COOH + NaOH --> CH₃COONa + H₂O

NaOH
C = 0.1 mol^.L^-1
V = 0.023 L
n = 2.3*10^-3 mol

Vinegar
n = 2.3*10^-3 mol
V = 0.025 L
C = 0.092 mol^.L^-1

C₁V₁ = C₂V₂
0.092 * 0.25 = C * 0.010
C = 2.3 mol^.L^-1

---OR---

D = 0.25 / 0.01
D = 25

D is the dilution factor. (0.01 L was diluted 25 fold to give 0.25 L)

G = 0.092 x 25
G = 2.3 mol^.L^-1

 

[fayad158]

thanks for that