北京大學(xué)定量分析化學(xué)簡明教程習(xí)題解答-第一章

1、 PAGE PAGE 4 第一章解：Ba(OH) +2HCl= BaCl +2H O222反應(yīng)前：nBa(OH) =0.150molL-10.0200L=0.00300mol2n(HCl)=0.150mo-10.0300L=0.00450mol反應(yīng)后：nBa(OH) =0.00300mol-(0.00450/2)mol=0.000750mol2cBa(OH) =0.000750mol0.0500L=0.0150molL-12m ( Zn )c(Zn 2 ) M ( Zn ) V0 .3250 g 0 .01988 mol L 165 .38 g mol 1 0 .2500 L解：解：n(1/2

2、)Na 2O = - n(1/5)KMnO 244 c(Na C O ) V(Na C O )= c(1/5)KMnO V(KMnO )22422444c(Na2O )= 0.050molL-124m(Na CO )= O ) V(NaO ) C O )2242222224=0.05molL-10.10L 134 gmol-1=0.67g解：C2+C2O2-=CaCO 42 CaCO+2+=HC+C2+2 42 2 45 HC+2MnO+6+=2M2+10CO+8HO2 2 4422)45 c(KMnO)V () M (CaO )5 0 .02 mol L 1 0 .030 mL 56 .08

3、 g mol 1m 244w (CaO) 2 0 .21g40 %解：Cr O O272m(Fe)= K Cr O ) V(K Cr ) M(Fe)/w(Fe)227227解：=(0.1200molL-11mL 55.85 gmol-1)/1%=0.6702gc (HCl) V (HCl) M (NaHPO12HO)w(HPO24O) 2 242m 0 .1012 mol L1 112 .0 %說明試劑失去結(jié)晶水NaHPO+HCl=NaHPO+NaCl 0 .02730 L 358 .0 g mol 10 .8835 g2424解：A3+H22=AlY+2H+(A3+)=(EDTA)(AlO2

4、 3)=(1/2)-n(EDTA)Z2+H22=Zn2-+2+-n(Zn2+)=-n(EDTA)w(11 c( EDTA) V ( EDTA ) c ( Zn ( Zn ) M ( Al O )O ) 22323m (0 .02081 mol L1 0 .03000 L 0 .02035 mol L1 0 .00650 L ) 102 .0 g mol 1 2 12 .43%0 .2018 g解：3.05mLNaOH相當(dāng)于 3.36mLHClCaCO+2HCl= CaCl +CO +H O3222n(CaCO)2 0 .1750 gc (HCl ) 3 0 .0954 mol L1V (HCl

5、)100 .1g 1 (40 .00 3.36 )mLc (NaOH) 0 .1053 mol L1HCl+NaOH=NaCl+ HO2解：H C H O +2NaOH= Na C H O +2H O244624462HCOOH+NaOH= HCOONa+HO2H C H O +10Ce4+2H +10Ce3+10H+244622HCOOH+2Ce4+=CO +2Ce3+2H+2Ce4+Fe2+=2Ce3+Fe3+ (H C H O ) 2446C H O V=c(Ce4+)V(Ce4+)- 2446解聯(lián)立方程得：c (H C H O )=0.03333 molL-12446解：(HCOOH)=0.0833mo-1(HC 2 H C42O =2C 2422-+3H O42cO H C O) c(NaOH (NaOH ) 0 .1500 mol L-1 25 .00 mL 0 .05000 mol L 12422O24C O )2243 25 .00 mL5HCO+2MnO-+6=2M2+10CO+8HO2 2 4422n H C O )V (KMnO4c() 54O