Question

AgCl(s)在纯水,0.01 mol82L-1 CaC12溶液,0.01 mol82L-1 NaCl溶液和0.05 m

Answer 1

AgCl的Ksp=1.8×10^-5,根据KSP不变,假设溶解了amol/L,有Ksp=c(Ag+)*c(Cl-)=1.8×10^-5 即纯水中 因而第二个Ksp=a1(a1+0.01x2)/1=1.8×10^-5,第三个Ksp=a2(a2+0.01)/1=1.8×10^-5第四个Ksp=a3(a3+0.05)/1=1.8×10^-5,没有计算器无法计算只能靠你自己计算了,得到的都是硝酸银溶解的浓度,