化学计算

Question

以0.02mol/L的EDTA 滴定浓度为0.02mol/L的pb2+、Ca2+混合溶液中的Pb2+，溶液的PH=5.0计算：（1）化学计量点时的lgk'(pbY)和[pb2+]、[Ca2+] （2）若以二甲酚橙作为指示剂，重点误差为多大？此时[Ca2+] 已知PH=5.0时，lgaY(H)=6.6,Ppb（终）=7.0（二甲酚橙） lgk(pbY)=18.0 lgk(CaY)=10.7

Answer 1

（1）lgk'(pbY)=lgk(pbY)-lgaY(H)=18.0-5.0=13.0
lgk'(CaY)=lgk(CaY)-lgaY(H)=10.7-5.0=5.7
[pb2+]=(0.01/10^13.0)^0.5=10^-7.5
[Ca2+]=(0.01/10^5.7)^0.5=10^-3.85
(2)Ppb(sp)=-lg[pb2+]=7.5
令x=Ppb(ep)-Ppb(sp)=7.0-7.5=-0.5
Et=(10^x-10^-x)/(c*K')^0.5=0.09%

Answer 2

4. (Y(Ca)=1+10-2+10.7=108.7 >(Y(H)
(Y≈108.7
(1). lgK'(PbY) = 18.0-8.7 = 9.3
pPb计=(9.3+2.0)/2 = 5.7 = pY'计=pCaY计[Pb2+]计=[CaY]计=10-5.7(mol/L)
(2). (pPb=7.0-5.7=1.3
101.3-10-1.3
Et = *100%=0.4%
10(9.3-2.0)/2
[PbY]10-2.0
[CaY]终=[Y']终= = =10-4.3(mol/L)
[Pb]K'PbY 10-7.0+9.3
或pCaY终=pY'终=pY'计+(pY = 5.7-1.3 = 4.4本回答被网友采纳

Answer 3

10(9.3-2.0)/2
[PbY]10-2.0
[CaY]终=[Y']终= = =10-4.3(mol/L)
[Pb]K'PbY 10-7.0+9.3
或pCaY终=pY'终=pY'计+(pY = 5.7-1.3 = 4.4