CN^- 12.73 mL 25.00 mL Ni^2+ Ni(CN)4^2- 10.15 mL 0.01307 M EDTA 39.35 mL

Cyanide solution (12.73 mL) was treated with 25.00 mL of Ni^2+ solution (containing excess Ni^2+) to convert the cyanide into tetracyanonickelate(II):

4CN^- + Ni^2+ → Ni(CN)4^2-

Excess Ni^2+ was then titrated with 10.15 mL of 0.01307 M EDTA. Ni(CN)4^2- does not react with EDTA. If 39.35 mL of EDTA were required to react with 30.10 mL of the original Ni^2+ solution, calculate the molarity of CN^- in the 12.73 mL sample.

Ni^2+ + EDTA → Ni(EDTA)^2-

---> Ni^2+ : EDTA = 1 : 1 계수비(= 몰수비)

39.35 mL of EDTA were required to react with 30.10 mL of the original Ni^2+ solution

(0.01307 mol/L) (39.35/1000 L) = 0.0005143 mol EDTA

---> original Ni^2+의 몰수

[original Ni^2+] = 0.0005143 mol / (30.10/1000 L)

= 0.017086 M

---> original Ni^2+의 몰농도

Cyanide solution (12.73 mL) was treated with 25.00 mL of Ni^2+ solution

(0.017086 mol/L) (25.00/1000 L) = 0.00042715 mol Ni^2+

---> 과량(excess) 가한 Ni^2+의 몰수

---> 이 양이 CN^-와도 반응(①)했고, EDTA와도 반응(②)했다.

Excess Ni^2+ was then titrated with 10.15 mL of 0.01307 M EDTA

(0.01307 mol/L) (10.15/1000 L) = 0.00013266 mol EDTA

---> EDTA와 반응한 Ni^2+의 몰수(②)

0.00042715 – 0.00013266 = 0.00029449 mol Ni^2+

---> CN^-와 반응한 Ni^2+의 몰수(①)

4CN^- + Ni^2+ → Ni(CN)4^2-

CN^- : Ni^2+ = 4 : 1 계수비(= 몰수비) 이므로,

4 × 0.00029449 = 0.00117796 mol CN^-

---> Ni^2+와 반응한 CN^-의 몰수

0.00117796 mol / (12.73/1000 L)

= 0.09253 M

---> CN^-의 몰농도

답: 0.09253 M

[키워드] 4CN^- + Ni^2+ → Ni(CN)4^2- 기준, 화학량론 기준, 정량분석 기준

CN^- 12.73 mL 25.00 mL Ni^2+ Ni(CN)4^2- 10.15 mL 0.01307 M EDTA 39.35 mL