- Tinh (H*), (OHl. pH cua dung djch
BAI TAP MAU
Bai 1. Tinh nong dp H \, pH ciia dung dich HCl 0.1 OM va dung dich NaOF COIOM. Giai • HCl — • H* + c r 0,10M 0,10M =>IH*] = 0,10M; pH = -lg[H*J = -lgO,l = l = > [ O H - ] = =1.0.10"'^M. 1.0.10"'
• NaOH > Na* + OH"
ỌOIOM ỌOIOM
=>[OH-] = 0.010M =>[H1= i:^:H!!l = 1.0.I0"'2M
1,0.10"^ =>pH = - l g l H * ] = - l g l 0 - ' ^ = 1 2 .
jPijilff 4$fi9 ¥^ phuong pWp 9i5i Hda hoc 11 VP CO - B5 Xuan Hung
Uhi Ị Dung dich axit axetic nong do 0,25M co pH = 2.
g) TInh do dien li cua axit axetic trong dung dich tren.
Neu hoa tan thtm 0,01 mol HCl vao 1 lit dung dich tren thi do dien l i cua
mit axetic tftng hay giam? GiU thich? Giai
a) PhiTctng trinh di6n l i : CH3COOH ^ CHiCOO" + H* Trong 1 lit dung dich c6 0,25 mol CH3COOH c6 pH = 2 Trong 1 lit dung dich c6 0,25 mol CH3COOH c6 pH = 2
=> [ H i = 10"''" = 10"^ = 0,01M
V|y troog 1 lit dung dich c6 0,01 mol CH3COOH phan li ra ion.
Bo di^n li cua CH3COOH la: a = — x 100% = 4%.
0,25
b) Khi them 0,01 mol HCl vao 1 lit dung dich tren thi nong do H"" tang len, do d6 cSn bhng dien li chuydn dich sang trai, do do do diSn U giam.
Bai 3. Mot dung dich c6 [ H ^ = 0,010M. Tinh [OH^ va pH cua dung dich. Moi tri/cfng cua dung dich nay la axit, kiem hay trung tinh? Hay cho biét mau cua quy tim trong dung dich naỷ
Gidi
[HI =0,010M => [0H-] = ' , =1,0.10-'^M
1,0.10"^ pH = - l g [ H l = - l g I 0 - ' = 2
Moi irirftng cija dung djch nay lii axil. Quy tim doi sang milu dọ Bai 4. Tinh pH cOa cac dung dich sau :
a) Dung djch KOH 0,003M. b) Dung djch H2SO4 0,02M.
c) Dung dich Că0H)2 0,004M (a = 0,8). d) Dung djch CH3COOH 0,004M (a = 0,7). d) Dung djch CH3COOH 0,004M (a = 0,7). Gidi a) KOH > K* + OH- 0,003M 0,003M [0H-] = 0.003M • I n 10"'"* ^ [ H i = i i ^ i i ^ = 0,33.10-'' : ^ pH = - l g [ H l = 11,48 3,0.10"^ b) H2SO4 > 2H^ + SÔ- 0,02M 0,04M [ H I = 0,04M => pH = - l g [ H l = 1,4. K H A W T f i Vim c) CăOH)2 > Cá^ + 20H- 0,004M 0,008M hay [OH-] = 0,008M
N6ng do OH" cua dung dich: [OH'] = 0,008.a = 0,008.0,8 = 0,0064M => [ H i = ^'"'^^ = l,5625.10-'2 =:> pH = - l g [ H l = 11,81.
6,4.10--^
d) CH3COOH -> CH3COO- + H^
0,004M 0,004M hay [H1 = 0,004M
=> Nong dp H* cua dung dich:
[ H i = 0,004.a = 0,004.0,7 = 0,0028M pH = - l g [ H l = 2,55. (adsbygoogle = window.adsbygoogle || []).push({});
Bai 5. Mot dung dich c6 pH = 9,0. Tinh nong do mol cua cac ion H"" va OH" trong dung dich. Hay cho bid't mau cua phenolphetalein trong dung djch naỵ
Gidi
lQ-14
pH = 9,0 [ H i = lO--^" = lO-^'M =>[0H1 = = lÓ^M
VI pH = 9,0 nen dung dich c6 moi trúdng kiem => phenolphetalein doi sang m^u hong.
Bai 6.
a) Cho hai chat NH3 va C6H5NH2 (anilin) chat nao c6 hang so bazd (kb) Idn
hdn? Giai thich?
b) Dung djch NH3 I M c6 a = 0,43%. Tinh hang so Kb va pH cua dung djch dọ
Gidi
a) V i phan tijT anilin c6 goc C6H5- la goc hut electron nen lam giam mat do
electron tren nguyen tijf N, do do c6 tinh bazcJ yeu hdn phan tiif NH3.
• ( N H 3 ) > ^ b ( C f i H s - N H a ) -
b) Phan tfng : NH3 + H2O NH^ + OH"
1 mol 0 0 ( l - x ) m o l xmol x mol Ma a = 0,43% = 0,0043 = y =:> x = 0,0043 = 4,3.10"^ T a c o : Kb = \^KllO^Kl±..jL= ^8,57.10-^- [NH3] 1-x 1-x 1-4,3.10"^ Vay Kb = 1,857.10"'.
Phan dsino va phiiong phip giai H6a hpc 11 VP c o - D5 XuSn Hung
IH*] = = J O : ! ! . = 0,23.10-'' = 2.3.10-'^
[ O H " ] 4,3.10"^
pH = - l g [ H l = -lg(2,3.10-'')= 11,64.
Bai 7. Tinh nong dp mol ciSa cdc ion H* va O H ' trong dung djch NaN02 1,0M.
Biet n^ng h^ng so phan li bazd cda NOj la Kh = 2,5.10'".
Giai
Phuang trinh di§n l i : NaNO: > Na* + N O j
I M I M Phan tfng thuy phan : N O j + H2O -> HNO2 + OH"
B a n d a u : I M 0 0 C a n b h n g : ( 1- x ) x x [ H N O , J . [ O H - l ^ ^ ^ [NO2] ( 1 - x ) „2 hay — = 2,5.10-" 1-x Giai ra ta c6: x = 5.10''^ hay [ O H ' ] = 5.10-''M [ H * ] = - l ^ i = - ^ ^ ^ = 0,2.10-''(M). 10-'^ _ 10-"* [ O H - ] 5.10"^
Bui 8. Cho dung djch X chua hon hpp gom CH3COOH 0,1M va CHjCOONa
0,lM. Biet a 25"C cua CH3COOH la 1,75.10-^ va bo qua sy phan li cua nuac. Tinh pH cua dung djch X.
'' Trich de thi tuyen sink Dgi hgc khoi B''
Giai
Phuang trinh dien li: CH.iCOONa CH3COO + Nâ (adsbygoogle = window.adsbygoogle || []).push({});
0,1 0,1 Can bang: CH3COOH -> CH3COO + Can bang: CH3COOH -> CH3COO +
Bandau: 0,1 0,1 0 Phan li: X x x Can bang: 0,1-x x + 0,1 x =:»Kc='^'^'^"^"'^^=: 1,75.10-^ ( D i e u k i | n: 0< x< 0, l ) 0,1-x => X ,= 1,75.10'(nh^n);X2 = -0,1 (lo^i) => pH = - Ig [ H = - log (1,75.10') = 4,76
Bai 9. Cho dung djch HCl c6 pH = 3. Ctin pha loang dung dich axit nay bhng
nurdc bao nhieu Ian dc thu diTtJc dung djch HCl c6 pH = 4?
Giai
Phtfdng irinh diOn l i : HCl > H* + Cl"
Goi V, la the lich dung djch HCl ban dau c6 pH = 3.
. pH = 3 IH*] = lÓ^M mh [H^] = - J l => n^^ ^ = lO-^V, (mol)
Goi V2 la the lich dung djch HCl sau khi pha loang c6 pH = 4. n +
. pH = 4 => [H*] = lO-'^M ma [H*l = - j ^ - => n^+ = 10-^V2 (mol) Khi pha loang dung djch so mol H* khong thay d o i : n + = n ^.
H d H s
=>10-^V, = 10-^V2 : ^ = ^ = 10 =:>V2=10V,
V$y pha loang dung djch HCl 10 Ian nghla la phai pha loang 1 the tich dung djch HCl vdi 9 the lich niTdc nguycn chát.
Bai 10. Dung djch X gom CH3COOH I M (Ka = 1,75.lÓ) va HCl 0.001 M . Tinh pH cAa dung djch X .
(Tn'ch de thi tuyen sink Dai hoc khoi A nam 2011)
Giai Ta c6: HCl - > H* + Cr 10-^ 10-^ CH3COOH < > CHjCOO + H* Bandau 1 0 l O ' Phanli x x x Can bhng 1 - x x x +10"^ T a c o : iLi^tlE!! = 1,75.10-^ (*) 1-x VI X < < 1 => 1 - X => 1, do do ttr (*) => x^ + 1 0 - \-5 X, = - 4 , 7 1 . 1 0 - ' ( l o a i ) X2 = 3.71.10' (nhan) => pH = - I g l H * ] = -lg(3.71.10-' + 10-') = 2,33
Bai 11. Trpn 100 ml dung djch hon hrtp gom H2SO4 0,05M va HCl ỌIM vdi 100 ml dung djch hon hdp gom NaOH 0.2M va BăOH)2 ỌIM ihu diTcJc dung djch X. Tinh pH cua dung djch X.
Phan d^ng phipng ph&p giii H6a hoc 11 VP co - D5 XuSn Hung
Gidi
Ta c6: n = 5.10'Wl; H H C I = 0,01 mol => 2 n + = 0,02 mol
H 2 S O ^ H
" B a ( O H)2 = 0.01 mol; HNaOH = 0,02 mol I n^^. = 0,04 mol Phan ung trung h6a: H* + OH" -* H2O Phan ung trung h6a: H* + OH" -* H2O
0,02 0,02 n . d„ = 0,04 - 0,02 = 0,02 mol n . d„ = 0,04 - 0,02 = 0,02 mol
O H
=> [OH] = 0,02 : 0,2 = 0,1M => pOH = 1 ^ pH = 13
Bai 12. Trpn Ian V ml dung djch NaOH 0,01M vdi V ml dung dich HCl 0,03M
difcJc 2V ml dung dich Ỵ Dung dich Y co pH la:
Ạ 4 B. 3 C. 2 D. 1
(Trich de thi tuyen sinh Ccio đn^ khoi A, B)
Gidi Ta C O : nNaOH = 0,0l.V mol Ta C O : nNaOH = 0,0l.V mol nHci = 0,03.V mol n =0,01.Vmol OH n ,-0,03.Vmol Phurong trinh ion: H^ + OH' H.O (adsbygoogle = window.adsbygoogle || []).push({});
0,01V 0,01V =^ = 0,03V - 0,01V = 0,02V (mol) =^ = 0,03V - 0,01V = 0,02V (mol)
H
H^ 0,02 V 2V = 0,01 = 10"2M =>pH = 2 ^ DapanC.
Bai 13. Tron 100 ml dung djch c6 pH = 1 gom HCl va HNO3 vdi 100 ml dung dich NaOH nong do a (mol/1) thu di/dc 200 ml dung dich co pH = 12. Gia Iri dich NaOH nong do a (mol/1) thu di/dc 200 ml dung dich co pH = 12. Gia Iri cua a la: (biel trong moi dung djch [H* ][0H""] = 10"'^)
Ạ 0,15 B.0,30 C. 0,03 D. 0,12.
(Trich de thi tuyen sinh Dai hoc khoi B)
Gidi
Ta co: pH = 1 :^ [ H ^ = IQ-'M =^ n ^ = 0,1.0,1 = 0,01 mol H
Va: V N a O H = 200 - 100 = 100 ml
fiNaOH = 0,1a mol n . = 0,1a mol
OH
Phifdng irinh phdn tfng: H* + OH" -> H2O H* + OH" -> H2O 0,01 0,01
Dung djch sau phan tfng co pH = 12 (moi triTdng bazcJ) => sau phan \ing tren
OH' dir, H* het.
I • Theo phiTdng irlnh phan tfog: n phaming = n . 0,01 mol
O H H
=> n _ j ^ = (0,1a-0,01) mol
O H
Mat kh^c, ta co: pH = 12 => pOH = 2 0,1a-0,01 0,2 = 0,01 a = 0,12 0,1a-0,01 0,2 = 0,01 a = 0,12
=> [OUT ] = 10-^ = ỌOIM
Dap an D.
Bai 14. Tron 250ml dung djch hon hdp HCl 0,08M va H2SO4 0,1M vdi 250ml
dung dich Bă0H)2 aM thi thu du'dc m gam kél lua va 500ml dung dich co
pH =12. Tinh a va m.
Gidi
Tacd: nH2S04 = 0,025 mol
"HCI - 0,02 mol
nBa(0H)2 = 0'25a mol
n ,=0,07 mol H+ H+ n 2 =0,025 mol •n = 0,5a mol OH" = 0,25a mol B a
Dung djch sau khi tron co pH = 12 (moi trifdng bazd) => Sau phan iJng OH" dif. => Sau phan iJng OH" dif. (adsbygoogle = window.adsbygoogle || []).push({});
H* + OH" ^ H2O 0,07 0,07 0,07 0,07 n,^„_ = (0,5a - 0,07) mol pH = 12 => pOH = 2 => [OH"] = 10"'M n . = 0,01.0,5 = 0,005 mol OH Ta co: 0,5a - 0,07 = 0,005 ^ a = 0,15 Bâ* + S04^- BaS04 i 0,025 0,025 mi = 0,025.233 = 5,825 (g) Dap an B. m BAITAPAPDyNG
Mi 1. Tinh nong dp mol cua ion H* trong dung dich HNO2 0,10M bie't ring hlng so phan li axit cua HNO2 la K a = 4,0.10-^ hlng so phan li axit cua HNO2 la K a = 4,0.10-^
Gidi
I PhiTdng trinh di^n li: HNO2 ^ H^ + NO2 Bandau: 0,1 OM 0 0 Bandau: 0,1 OM 0 0 Canbkng: (0,1-x) x x
Phan djinfl v i phuang pMp giii H6a hqc 11 VP cO - Dg Xuin Hung
[HNO2] (0,1- X ) 0,1- X
Vi HNO2 la axit ycu nen x « 0,1 nen — = 4.10"* => x = 6,3.10"^
0,1
Vay [H1 = 6,3.10-^M. Bai 2. Co hai dung dich sau :
a) CHjCOOH ỌIOM (K, = 1,75.10"*). Tinh nong do mol cija ion H\
b) NH3 0,1 OM (Kh = 1,80. lO-'). Tinh nong do mol cua ion OH".
Giai a) CH3COOH CH,COO- + H* Bandau: 0,1M 0 0 Canb^ng: 0.1-x x x ^ _ [CH3COO-].[H-] X.X _ x^ 5 [CH3COOH] 0 , 1- X 0,1- X ' •
Vi CH3COOH la mOt axit yeu nen X « 0,1 nen — = 1,75.10"^ 0.1 Giai r a t a c 6 x = 1,32.10-^M hay [H*] = 1,32.10'^M. b) NH3 + H2O -> N H ; + OH- Bandau: 0.1 0 0 (M) Canbkng: 0,1-x x x (M) [ N H ; ] . [ 0 H- ] _ X.X _ x^ 3 [NH3] - 0 , l - x - 0 , l - x - ^ ' ^ ° - ^ ° VI NH3 la mpt bazd yeu nen x « 0,1
— = 1,80.10"^ =>x= 1,34.10"^ Vay [0H-] = 1,34.10-^M.
Bai 3.
a) Them tir tir lOOg dung dich H2SO4 98% v^o niTdc va dieu chinh de diTcJc 1 lit dung dich Ạ Tinh nong do mol cua ion H* trong dung dich Ạ dung dich Ạ Tinh nong do mol cua ion H* trong dung dich Ạ
b) Phai th6m v^o 1 lit dung dich A tren bao nhieu lit dung dich NaOH 1,8M de thu dÚdc: thu dÚdc:
- Dung djch c6 pH = 1; - Dung dich c6 pH = 13.