大学无机化学课后习题解析第三章

《大学无机化学课后习题解析第三章》由会员上传分享，免费在线阅读，更多相关内容在工程资料-天天文库。

1、第3章酸碱反应和沉淀反应习题参考答案解：(1)pH=-lgc(H+)=12.00(2)0.050molL_1HOAc溶液中,HOAcH++OAc'cW(mol*L_l)0.050-xxxKg=c(H+)c(OAc-)=x-x=18x1Q-5c(HOAc)0.05()-xc(H*)=9.5xlO-4molL'1pH=-lgc(H+)=3.022.解：(l)pH=1.00c(H+)=0.10molL-1pH=2.00c(H+)=O.OlOmolL'1等体积混合后：c(H+)=(O.lOmolL^+O.OlOmolL1)/2=0.055mol-L'1pH=-l

2、gc(H+)=1.26(2)pH=2.00c(H+)=0.010molL'1pH=13.00pOH=14.00-13.00=1.00,c(OH)=O.lOmolL'1等体积混合后：c(H+)=O,O1Q1^01L*=0.0050mol-L-*O.lOmolL1nncniT.1c(OH)==0.050mol・L酸碱中和后：H"+OH'fH2Oc(OH)=0.045molL_,pH=12.653.解：正常状态时pH=7.35c(H+)=4.5xlO-8molL_1pH=7.45c(H+)=3.5xl0-8molL_1患病时pH=5.90c(H+)=1.2x1

3、0-6molL_11.2x10'6mol•L'1“8=274.5x10%mol-L'11.2xl()-%wl-L-13.5xIO"8mol・L“患此种疾病的人血液中c(Hj)为正常状态的27〜34倍。假设在l.OLO」0mol・L“氨水中加入xmolNH4C1(s)oNH3H2ONH/+OH"0.10-1.Ox10'5x+l.OxlO'5c平/(mol-L'1)LOxlO'5+—ONI—)C(Oll)c(nh3-h2o)(NH3・匕0)(x+1.0xl0-5)i.0xl(r'=18x10-5O.lO-l.OxlO-5x=0.18应加入NH4CI固体的质量

4、为：0.18molL"xiLx53.5gmoP=9.6g6.解：设解离产生的H+浓度为xmol-L'1,则HOAcH*0.078-x+OAc'0.74+xC(HOAc)0.74x一5z而厂I1.9X10JpH=-Igc(H+)=5.72向此溶液通入O.lOmolHCl气体，则发生如下反应：NaOAc+HC1-NaCl+HOAc反应后：c(HOAc)=0.18mol-L1,c(OAc")=0.64mol-L1设产生的Ff变为x^ol-L'1,则HOAc二0.18・x'(0.64+x')x‘0.18—x，H++OAc'x‘0・64+x‘lFxlO"x,=5.

5、1xio6,pH=5.30A(pH)=5.30-5.72=-0.427.解：(1)设NH4CI水解产生的H十为xmol-L1,则NH3H2O+ITc平/(molL")NH/+H2O0.010-xKO_c(NH3H2O)c(Hj_c(NH：)^f(NHrH2O)=5・6xl0"°4.解：一元弱酸HA,pH=2.77c(H+)=1.7xl0-3molL_1HAH*+A'cW(molL')0.10-1.7X10'31.7x10'31.7"0以K&=c(H+)c(AJ=(1.7x10-3)2"一c(HA)~0.10-1.7X10-3=2.9xl0'5◎=5.6x

6、10」°0.010-Xx=2.4xl0-6,pH=5.62(2)设NaCN水解生成的疋为x'molL",则CN"+H2OHCN1.7x10-3(TToxl00%=1.7%5.解：溶液的pH=9.00,c(H+)=1.0x10-9molL_1故c(OHJ=l.OxlO^molL'1OH'c平/(moll?)0.1OXc(HCN)c(OH-)_显c(CN-)hx'=1.3x10-3,pH=11.118.解:(1)K,/(HC1O)=2.9x1O'8；(2)Ksp°(AgI)=8.51xl0'179.解：(1)设CaF2在纯水中的溶解度G)为xmolf'o因为

7、CaF2为难溶强电解质，且基本上不水解，饱和溶液中：所以在CaF210.解：(1)由题意可知：c(Mg2+)=0.050mol-L'1当c(M『+).{c(OH)}2>Ksp°(Mg(OH)2)时开始有Mg(OH)2沉淀出。CaF2(s)Ca2++2F2xc平/(moll?)心(Mg(OH)2)c(M/+)/5.61xl0~12V5.0X10-2=l.OxlO-'molL

8、1Ca2++2F2y+1.0x1O'2(2){c(A严)}•{c(OH)}3=4