本科仪器分析课件-仪器分析习题解答

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1、董慧壮编2005年2月第二章电化学分析法习题解答25.解:pHs=4.00、pHx=pHs+Es=0.209VEx-Es0.059(l)pHx1=4.00+0.312-0.209~~0.059=5.75(2)pHx2=4.00+0.088-0.2090.059=1.95(3)pHx3=4.00+-0.017-0.2090.059=0.1726.解：[HA]=0.01mol/L?E=0.518V[A-]=0.01mol/L?①SCE=0.2438VE=①SCE—①2H+/H20.518=0.2438—0.059lg[H+][H+]=0.010^0

2、10.518=0.2438—0.0590.010^01^ka4.647ka=2.25x10」27・解：2Ag++CrO亍=Ag2CrO4[Ag+]2=Ksp[CrOME—0SCE—©AgQO訂Ag・0.285=0.2438-[0.799+0.059Ksp2~T~8[CrO；-]為—-,0[CrO^]=l.lxlO'126.93xlO~10-3=1.59x10(mol/L)28・解：pBr=3,-3aBr・=10mol/Lpci=1,-1aci_=10mol/L百分误差=KBr-,crxacrXi。。—6x10十10“灯00—60IO'3因为干扰

3、离子C「的存在，使测定的aB.变为：-3,-3-1-3a_=a_+Kd…xa_=10+6X10X10=1.6X10BrBrBr.ClCl即■由10_3mol/L变为1.6X10_3mol/LBr相差3.0-2.8=0.2pBr单位29.解:%01如1pHpH/V0.002.901.003.010.112.003.150.143.003.340.194.003.570.235.003.800.236・004・030.237.004・340.318.004.810.478.405.251.108.605.611.308.806・202.959.00

4、6.803.009.209.1011.509.409.803.509.6010.151.759.8010.411.3010.0010.711.50(1)由pH-V图查得V终点=9」0ml由厶pH/AV-V图查得V终点=9」0ml(2)计算[HA][HA]=0.1000x9.10_Fooo=0.09100(mol/L)(3)化学计量点的pH值pH6.809.10A2pH/AV2+42.5-40.0pH终占=6.80+9.1-6.842.5+40.0x42.5=7.9830.解：R=1二=eGI1e20.0R]=317(Q)Gi&0.06301e2

5、0.0r2=26.7(Q)G2k20.750所对应的电阻值范围是26.7-317Qo31・解：根据所给电阻值，求出电导值，具体如下:V/mlG/S0.003.15xlO'41.002.60xlO'42.002.04xlO'43.001.54xlO'44.001.97xlO'45.002.86xlO"46.003.66xlO'4以电导值为纵坐标加入的HC1体积为橫坐标作图。从右上图可得化学计量点时HC1的体积为3.25ml3.25X1.00=CxX100Cx=0.0325(mol/L)即溶液中NaOH的浓度为0.0325mol/L。32・解：(1

6、)①°严心=叭皿+警T=・0.403+^^lg[0.0167]=■0.455(V)①AgCHAg=①AQAg一0.059lg[Cl_]=0.222—0.059lg(0.0334)=0.309(V)E池=0.309—(-0.455)=0.764(V)(2)E外=0.764+iR=0.764+0.0283X6.42=0.946(V)33・解：(1)J①S讪⑷=0-71>①二对*=-0.152且[Br_]=[r]=0.0500mol/L显然1一应在电极上先沉积，BL开始沉积的电极电位为：①2=①。t—0.059lg[Br-]=0.71一0.059lg

7、[0.0500]=0.148(V)[K]=1.00X10~6mol/L时的电极电位为：①1=①八如仪=①°A和A*-0.059lg[K]=-0.152一0.059lg[1.0X10^]=0.202(V)对于阳极过程①I必须负于(小于)①2方可定量分离，而今①I>①2,故不能定量分离。(2)因为①,,gw=0.222>①。切⑷=-0.152故I-先在电极上沉积，C1•开始沉积的电位为：①2=AgCl/Ag="AgCl/Ag-0.059lg[C「]=0.222—0.059lg[0.0400]=0.304(V)由⑴已求得®i=0.202V,在①1(0

8、.202V)和①2(0.304V)间任选一电位电解，「和CT均可定量分离。⑶由上述计算可知I一和CT口J定量分离，电极电位(相对于NHE)在①1=0.