+ OH" >H20
0,03 0,03
=> n^^ = 0,04-0,03 = 0,01 mol ^ [Hi = I O ' M => pH = 1.
Bai 2: Hoa tan hoan toan 8,94 gam hon hdp gom Na, K va Ba vao niMc, thu
dMdc dung dich X va 2,688 lit khi H2 (dktc). Dung dich Y gom HCl va H2SO4,
ti le mol tu-Ong ij-ng la 4 : 1 . Trung hoa dung djch X bdi dung dich Y, tdng
khói liTtJng cac muoi di/cfc tao ra la
Ạ 13,70 gam. B. 18,46 gam. C. 12,78 gam. D. 14,62 gam.
Gidi
Tac6: nH2 = 0,12 mol
Sđophanurng: H2O > OH" + - H 2
2
0,24 0,12 Goi nH2S04 = ^ '"ol -> HHCI = 4x mol -> = 6x mol Goi nH2S04 = ^ '"ol -> HHCI = 4x mol -> = 6x mol
Phan u-ng trung hoa: + OH^ -> H2O
0,24 0,24 6x = 0,24 => X = 0,04 mol 6x = 0,24 => X = 0,04 mol
=> nH2S04 = 0,04 mol; UHCI = 4.0,04 = 0,16 mol
Ta c6: Khói-lu'dng muoi = khói liTdng kim loai + khói liTpng góc axit
-> Khói liTcJng muoi = 8,94 + 0,04.96 + 0,16.35,5 = 18,46g Dap an B. Bai 3: Cho dung dich chtira 0,1 mol (NH4)2C03 tac dung vdi dung djch chiJa 34,2 Bai 3: Cho dung dich chtira 0,1 mol (NH4)2C03 tac dung vdi dung djch chiJa 34,2
gam Bă0H)2. Sau phan u-ng thu difdc m gam kct tuạ Tinh m.
Gidi
Ta c6: n(NH4)2C03 = O-lmol => n^^^j- = 0,1 mol
, nBăOH)2 = 0,2mol n^^2+ = ^,2 mol Pu: Bâ^ + COj^- ^ BaCOj i Pu: Bâ^ + COj^- ^ BaCOj i
0,1 0,1
Vay: mi = 197.0,1 = 19,7 (gam)
40 »
Bai 4: Nho tCr tCr 0,25 lit dung dich NaOH 1,04M \ko dung dich gom 0,024 mol
FeCls; 0,016 mol Al2(S04)3 vk 0,04 mol H2SO4 thu dUOc m gam ket tuạ Tinh m.
Gidi
Taoo: UNaOH = 0,26 mol => n . = 0,26 mol
OH
"FeCis = 0'024mol ^ Up 3^. = 0,024 mol nAi2(S04).-, = 0,016mol n^^,^ = 0,032 mol nAi2(S04).-, = 0,016mol n^^,^ = 0,032 mol n = 0,04mol => n . = 0,08mol (adsbygoogle = window.adsbygoogle || []).push({});
H2SO4 H+
Khi cho đ NaOH vao đ gom FeCls; Al2(S04)3; H2SO4 ; cac phan I'mg Ian li/dt xay ra : li/dt xay ra : H^ + OH" -> H2O 0,08 0,08 Fê*+ 30H" -> Fe(OH)3 0,024 0,072 0,024 Al^^ + 30H- -> Al(OH)3 0,032 0,096 0,032 I n . = 0,08 + 0,072 + 0,096 = 0,248 mol OH => n .j„ = 0,26-0,248 = 0,012 mol OH
Do OH" du nen tiep tuc xay ra phan ung hoa tan ket tua A1(0H)3. A1{0H)3 + OH" -> AIO2" + 2H2O A1{0H)3 + OH" -> AIO2" + 2H2O
0,012 0,012
=> so mol Al(OH)3c6ni,i = 0,032 - 0,012 = 0,02(mol)
Vay: m i = mA,(OH)3 c6n„i + mp,(OH)3 = 0,02.78 + 0,024.107 = 4,128(g)
Bai 5: Cho 200ml dung dich A chiJa HCl IM vil H N O 3 2M tac dung vdi 300ml dung dich chuTa NaOH 0,8M va KOH (chifa bict nong do) thi thu du^dc dung dung dich chuTa NaOH 0,8M va KOH (chifa bict nong do) thi thu du^dc dung dich C. Biet rhng de trung hoa dung dich C can 60ml HCl IM. Tinh nong do mol ciia dung dich KOH.
Gidi
Gpi x M la nong do mol cua dung dich KOH: Taco: nNaOH = 0,24 mol Taco: nNaOH = 0,24 mol
" K O H = ^'•^^ "^ó nHN03 = 0,4 m o nHN03 = 0,4 m o " H C I 0,2 mol => Y n = (0,3x + 0,24)^ OH" mol nHN03 = 0.4 mol Tn + = 0,6 H+ mol
PhSn dgng vci phudng ph^p giSi H6a hpc 1 1 VP cO - D 5 Xufln Hi;ng w
Khi tron iSn hon hrfp 2 axit vil 2 bazcJ xay ra phan i^ng triing h6a:
+ OH- ^ H2O (1)
0,6 0,6
Sail phan iJng (1) can dung HCl de trung hoa, chifng to OH" dir.
=> n . = (0,3x + 0,24) - 0,6 = (0,3x - 0,36) mol
OH
Ta c6: U H C I = 0,06 mol => n^. = 0,06 mol
H
Phiin xing trung hoa giffa HCl vii dung djch C: H^ + OH" -> H2O
0,06 (0,3x - 0,36) => 0,3x - 0,36 = 0,06 => x = l , 4 M
Bai 6: Cho dung djch Bă0H)2 den dU vao 50ml dung dich X c6 chiJa cac ion NH4''; SO4'" ; NO," thl c6 11,65 gam k c l lua dmc tao ra va dun nong c6 4,48 lit khi bay ra (dktc). Tinh nong do mol moi muói trong dung djch X.
Gidi Ta c6: n^^.^ = - ^ - ^ = 0,05 m o l ; n^ H , = — = 0,2 mol Bab04 ' 22 4 Bá* + S04^- -> BaS04>l' 0,05 0,05 NH4* + OH" NH3 + H2O 0,2 0,2 Dung dich X c6 cdc ion NH4^; 504^"; NO,r
CO 2 muoi la (NH4)2S04 va NH4NO3. Ta c6: n,^^^,^,^, - ^of = ^ ' ^ ^ "^"^ =^ [ ( N H J . S O , nNH4No,, = 0,2 - 0,05.2 = 0,1 mol => [ N H ^ N O J 0,05 0,05 0,1 0,05 = 1M = 2 M
Bai 7: Dung djch A chiJa HCl I M va H2SO4 0,6M. Cho 100ml dung djch B gom KOH I M va NaOH 0,8M vao 100ml dung djch A, c6 ciin dung djch sau phan úng thu diWc m(g) chift r^Cn. Tinh m.
Gidi
Dung dich A c6: n^^^ô = 0,06 mol •
" H c i ^ O . l m o l
Dung dich B c6: n^^oH = 0,08 mol
n^, = 0,22 mol (adsbygoogle = window.adsbygoogle || []).push({});
H
" K O H = 0,1 mol n^^. = 0,18 mol OH
Dd sau phan úng gom H ' + OH" ^ H2O 0,18 0,18 • H* dif: 0,22 - 0,18 = 0,04 mol : 0,1 mol Nâ : 0,08 mol cr : 0,1 mol • SO4- : 0,06 mol
Khi CO can dung djch HCl bay h d . Vay khói lifdng chat ran thu difdc c6 giii tri: m = m,„,ị,„ + m,„ị,n = 0,1.39 + 0,08.23 + 0,06.35,5 + 0,06.96 = 13,63 (g)
Dana 5. Bai t^p ap dung djnh lugt bao loan di^n tich. BAI TAP MAU VA BAI TAP AP Dl^NG BAI TAP MAU VA BAI TAP AP Dl^NG
Bai 1. Mot dung djch chuTa 0,02 mol Cu^*, 0,03 mol K*, x mol CI" va y mol SÔ". Tong khoi lifdng ciic muói tan c6 trong dung dich la 5,435 gam. Tim x va ỵ
Gidi
Khi tron dung dich A \di dung dich B xay ra phan Ofng:
/^p dung djnh luat bao loan dien tich, ta c6:
X + 2y = 0,02.2 + 0,03 = 0,07 (1)
Ta lai c6: mmu,-,: = Sm,„,ị.n + Ima„K,n
=> 35,5x + 96y = 5,435 - (0,02.64 + 0,03.39) = 2,985 (2)
T i r( l ) v a ( 2 ) = > ^ = ° ' ' ' [y = 0,02
Bai 2. Mot dung dich c6 chtfa cac ion: x mol 0,2 mol M g ^ \3 mol Cu^^ 0,6 mol S04^", 0,4 mol NOf. Co can dung dich nay thu dúrtc 116,8 gam hon hcJp cac muói khan. Tim M .
Gidi
Ap dung dinh luat bao loan dien tich, ta c6: 3x + 0,2.2 + 0,3.2 = 0,6.2 + 0,4 x = 0,2 mol Ta c6: m^^,;= m ^ . , + n i ^ ^ , +m^^,. +m^^,^ +m^^_
=> 116,8 = 0 , 2 .M M + 0,2.24 + 0,3.64 + 0,6.96 + 0,4.62
=> M M = 52 => M la Cr
^ai 3. Dung djch A c6 chiJa 5 ion: Mg^^ B a ' \\1 mol CI" va 0,2 mol NO3'.
Them dan V lit dung dich NajCOj I M vao A den khi du^dc lifdng ket tua Idn nhat. Tim V.
Phan dgng vh phuong ph^p giSi H63 hoc 11 V6 cd - D5 XuSn Hung
Gidi
PhifcJng trinh ion rut gon:
Mg'* + CÔ- > MgCOjl Bâ*+ CO3- > BaCOji Bâ*+ CO3- > BaCOji Câ* + CO3" > CaCOjJ-
Khi phan ỉng két thilc, cac két tua tach khoi dung dich, phan dung dich chiJa
Na"", cr va NO3". De trung hoa dien thi: n^^^^ = n^^. = 0,3 mol
=> V,,,,^co, = ^ ^ = °;^=0,15(l)=150ml I Na 2
Bai 4. Dung dich A chiJa cac ion COj^", SOj^", SÔ^', 0,1 mol HCO3" va 0,3 mol Nâ Them V lit dung dich Bă0H)2 IM vao dung dich A thi thu diTdc Ming Nâ Them V lit dung dich Bă0H)2 IM vao dung dich A thi thu diTdc Ming
ket tua Idn nhat. Tim V.
Giai (adsbygoogle = window.adsbygoogle || []).push({});
Nong do cac ion [Bâ^] = IM; [OH"] = 2M. De thu di/cJc liTdng ket tua Idn nhat, can 0,1 mol OH" de tac dung het vdi HCO3". nhat, can 0,1 mol OH" de tac dung het vdi HCO3".
HCO3' +0H" ->C03^"+ H2O
Mat khac, can 0,3 mol OH' de trung hoa Na"".
Vay tdng so mol OH' can la 0,1 + 0,3 = 0,4 mol => ny^joH,^ = 0,2 mol The tich dung dich Bă0H)2 la: V = ^ = 0,2 lit The tich dung dich Bă0H)2 la: V = ^ = 0,2 lit
I
Bai 5. Dung dich X chtfa cac ion: Fé\, NH4. Chia dung dich X
thanh hai phan bang nhau:
- Phan mot tac dung vdi liTcJng du" NaOH, dun nong thu di/dc 0,672 lit khi d (dktc) va 1,07 gam ket tuạ (dktc) va 1,07 gam ket tuạ
- Phan hai tac dung vdi liTdng dúdung dich BaCl2, thu dúdc 4,66 gam két tuạ Co can dung djch X thu diTdc m gam muói khan (qua trinh c6 can chi cd nÚdc