分析化学第五章配位滴定法习题答案

1.解:∵fMnYfMnYYHKK()()()lglglg8∴YH()lg13.87-8=5.87查表可得:pH5.22.解:⑴查表得:YH6.45()10⑵∵YHCYCY6.45()()10()∴cYcY96.45()()7.1010103.解:滴定50%时,cCamolLmolL2310.000.01()/3.3310/20.0010.00∴pCa=2.48滴定100%时,Ca2与EDTA恰好完全反应,此时,可认为溶液中的Ca2完全来自CaY的解离.∵fCaYfCaYYHKK()()()lglglg10.690.4510.24∴fcCaYcCaYKCaY7()()/()7.5910∴pCa=6.12滴定200%时,cYmolLmolL320.000.01()/3.3310/20.0040.00fCaYcCaYKcCacY()2()10.24()()fcCaYcCamolLKCaYcY21110.243()0.01020.00/(20.0040.00)()5.7510/()()103.3310∴pCa=10.244.解:∵nZnClmolLLmol234()0.01032/20.56102.1210∴mZnClmolgmol24()2.1210126/0.02673∴质量分数0.10695.解:∵nMgNHPOmolLLmol3444()0.01000/20.00102.0010∴mPmolgmolg43()2.001031/6.210∴磷的含量为P3()(6.210/0.1000)100%6.2%6.解:mMgmolLLgmolg233()0.05000/4.101024/4.9210mZnmolLLgmolg23()0.05000/13.401065/0.04355mCumolLLgmolg23()0.05000/(37.3013.40)1064/0.07648所以，Cu、Zn、Mg的质量分数依次为：61.18%、34.84%、3.94%7.解:fZnYfZnYYHKK()()()lglglg16.506.4510.05∵fZnYK()lg0.016∴可以准确滴定8.解:⑴pH=6.0时,fCaYfCaYYHKK()()()lglglg10.694.566.13∵fCaYKcCa2()lg()6∴此时不可以准确滴定⑵如果想准确滴定,则:fCaYfCaYYHKK()()()lglglg8∴YH()lg2.69∴pH7.69.∵pH=3时,fCuYfCuYYHKK()()()lglglg18.8010.608.20∴可以准确滴定同理,pH=5时,也可以准确滴定∵spKCuOHcOHcCu22202[()]()()2.2010∴spcOHKCuOHcCu292()[()]/()1.4810∴pOH=8.23,pH=5.2又∵fCuYKcCu2()lg()6fCuYfCuYYHKK()()()lglglg8∴YH()lg18.80-8=10.80∴pH3.0所以，滴定的范围在pH=3.0～5.210解：在50ml试液中：∵nCdmolmol234()0.021746.76101.4710∴mCdg()0.01650nBimolmol334()24.67100.024796.1210mBig()0.01650nPbmolmolmol2334()0.021746.76100.021746.76104.8910mPbg()0.1012∴Pb、Cd、Bi质量分数依次为：20.91%，3.41%，26.43%。11.解:nZnmolLmol233()22.30100.1000/2.2310所以：mZngmolmolg3()65/2.23100.1450Zn的质量分数为：36.25%