14
Giai
a) Khoi liTdng chat tan H2SO4 :
C%.m,,d 98%. 100 , 98 , . ,.
'"^••= - ^ = -lB5^ = ''^^^'"^=^ " " - - ^ = ^ - U n . o l )
H2SO4 > 2H* + S O ^
1 mol 2 mol
=>[H1= Y = 2(M).
b) Goi V la the tich dung djch NaOH c6 nong do 1,8M
= > n N a O H = 1.8.V(mol)
2NaOH + H2SO4 )• Na2S04 + 2H2O 1,8V 0,9V = > n H 2 S 0 4 d ^ = ( l - 0 . 9 V ) mol Sau phdn ỉng : Vjj = 1 + V . p H = l =>[H1 = 10-'=0,1(M) H2SO4 > 2H* + S O ^ (1-0,9V) (2-1,8 V) = > [ H 1 = ^ - ^ = 0.1 1 +V Giai ra ta CO V = 1 (lit).
• pH= 13 => dung dich c6 tinh bazd
=> [H*] = lO-'^M => [0H-] = 10-'M = 0,1M
H2SO4 + 2NaOH > Na2S04 + 2H2O
I mol 2 mol
= > n N a O H d i f=(1.8V-2)moI NaOH > Na* + OH"
(l,8V-2)mol (1,8V-2) mol 1 8 V - 2
=> [0H-] = ' = 0.1 V = 1.2353 (lit). 1 +V
Bai 4. Tinh the tich dung dich Bă0H)2 0,025M can cho vao 100ml dung djch hSn hdp gom HNO3 v^ HCl c6 pH = 1 de tao th^nh dung dich c6 pH = 2.
Giai
• D u n g djch hon hcJp gom HNO3 v^ HCl c6 :
l p H = 1 =>[H1 = 10"' =0.1Mii> n , =0,1.0,1 = 0,01 (mol)
[Goi V \h. th6 tich dung dich BăOH)2 0,025M ^
Phati djing va phuang phap glSi H6a hge 11 VP c o - D5 Xuan Hi/ng
nBăOH)2 = 0,025.V (mol) BăOH)2 > Bâ*+ lOW
0,025V 0,025V (mol)
Phifdng trlnh phin tfng : H* + OH" — » H^O 0,05V 0,05V
Dung dich tao thanh c6 pH = 2 => [ H i dir = 10"''" = 10-' = 0,01M Vjj sau phan iJng = V + 0,1
^ V d > / = ^ M - V = 0,01.(V4-0,l)
oO,01 - 0,05V = 0,01V + 0,001 o V = 0,15 (lit). Bai 5.
a) Tinh pH cua dung dich chtfa l,46g HCl trong 400ml.
b) Tinh pH cua dung djch tao thanh sau khi tron 100ml dung dich HCl 1,00M vdi 400ml dung dich NaOH 0,375M.
Gidi 1 46 a) nHci= — - = 0,04 (mol) 36,5 HCl — > H* + c r 0,04 0,04 (mol) ^ [ H I =^ = ^ = 0V 0,4 ,im=>pli = -IgO, 1 = 1. b) Ta CO : nNaOH = 0,4.0,375 = 0,15 (mol)
NaOH > Na* + OH" 0,15 mol 0,15 mol
nHci = 0,1.1 =0,1 (mol)
HCl — > H^ + c r
0,1 mol 0,1 mol
Vdj sau phan ỉng = 0,4 + 0,1 = 0,5 (lit).
PhiTcJng trmh phan tfng : H^ + OH" — - » H2O
0,1 mol 0,1 mol n^..- . =0,15-0,1 = 0,05(mol)
"OH dU
[OH-] = - ^ = 0,l(M>
[Hi = h2:lI^=x(r^^.(Pi^-^ 0^110^^ = [§,
Bai 6. Tinh nong do [H*] va [OH"] trong dung dich CH3COOH 2M. Neu sau do
them v.^o moi lit dung djch axit tren 0,2 mol muói CHjCOONa thi [H*] va [OH"] cua dung dich tang hay giam bao nhieu Ian. Bic't hang só phan 11 axit la 2.10"'* va the tich dUng dich thay dói khong dang kẹ
Gidi
Phircfng trinh dien l i : CH3COOH -> CH^COO" + H* (1)
Bandau: 2 0 0 (M) Can bang: 2 - x x x (M)
K . = [CH3C00-j.[H"]^ x.x ^ ^2 10-5
[CH3COOH] 2 - x 2 - x Vi la axit yeu nen x « 2.
2 = ^ ^ = 2.10-5 = ^ ^ = 2.10-5 2 X = 6,324.10" => [ H i = 6,324.10'^ {M)=> [0H-] = 10 -14 = 1,58.10"'^ (M). (2) 6,324.10"
Neu them vao moi lit dung djch axit 0,2 mol muoi CHsCOONa : Phi/dng trinh dien l i : CHjCOONa > CHjCOO" + Na*
0,2 mol// 0,2 mol//
nen lam cho nong do CH3COO- trong dung djch tang len do do can bang d
phdn tfng ( i ) dich chuyen theo chieu tOf phai sang irai, nen can b^ng duftJc thanh lap. CH3COOH -> CH3COO- + H* Bandau: 2 0 0 Canb^ng: ( 2 - y ) (0,2+ y) y Vi Ih axit yeu ndn y « 2 => [CH3COOH] = 2 - y = 2 (M) [CH3COO-] = 0,2 + y = 0,2 (M) t C H , C O O - l . [ H ' 1 ^ 0 ^ ^ ^ , ^ . . [CH3COOH] 2 => y = 2.10"* hay [ H i = 2. lO"" (M) (M) (M) [OH1 = 10 -14 = 0,5.10"'" =5.10-" (M) 2.10 -4
So vdi dung dich CH3COOH ban diu:
[ 0 H 1 tang len = 5.10 - 3 r W e- 4 i^ 4 f t- v 4 - t t I 1 gi^ni 31,64 Ian.
l.:;aH).7'ViEN TINH BIN.H THUAN '
PhSn dgng va phuong phap giii H6a hgc 11 Vfl cO - B5 XuSn Hang
Bai 7.
a) Tinh pH ciia dung dich A la hon hcJp gom HF 0,1M va NaF = 0,1M. b) Tinh pH cua 1 lit dung djch A d tren trong hai trUdng hcTp sau : - Them 0,01 mol HCl vao;
- Them 0,01 mol NaOH vaọ
Biét rang hang so axit (hhng so ion hoa) cua HF la = 6,8.10""*. Cho log6,8 = 0.83.
Gidi
a) Phtfdng trinh dien 11: HF - > H^ + F
Trong dung dich c6 F" nen lam cho can bang it bi chuyen dich nen c6 the coi [HF] = 0 , 1 M = > [ F ] = 0,1M k^=mlLlE:ị6,8.10-^M) [HF] [H1=^'^-^Q"'-^'^=6.8.10-"(M) 0,1 pH = - l g [ H l = -lg(6,8.10"') = 3,17.
b) * Khi them 0,01 mol HCl vao t h i t a c 6 : H ^ + F > HF [HFJ = 0,1 + 0,01 = 0,11 (mol)
[ F ] = 0 , 1 - 0 , 0 1 =0,09 (mol)
[ H I = 6,8.10-r = 6.8.10"^ — - 8,31.10-^ [ F - ] , 0,09
=> pH =-lg(8,31.10"") = 3,08.
* Khi them 0,01 mol NaOH vao ta c6 : HF + OH" > F + H2O
=:>[F] = 0,1+0,01 =0,11 (mol) [ H F ] = 0 , 1 - 0 , 0 1 =0,09 (mol)
[H^] = 6.8.10-*. = 5,56.10"^(M) pH = - l g [ H l = 3,25.
M i 8. Tron ba dung dich H2SO4 0,05M; HNO3 0,1M va HCl 0,15M vdi nhffng the tich bang nhau, thu dtfcfc dung dich Ạ Lay 600ml dung djch A cho tac dung vcti dung dich B gom NaOH 0,3M va KOH 0,15M. Tinh the tich dung dich B can dijng de sau khi tac dung vdi 600ml dung dich A thu difdc dung dich C O pH = 3.
Gidi
V i the tich b l n g nhau nen the tich moi dung dich axit trtfdc khi trpn Ian la: — = 200 ml = 0,2 (lit)
Sómol m6i axit trong 600ml dung dich A : . . . .
nH2S04 =0,05.0,2 = 0.01 (mol)
nHN03 =0,1.0,2 = 0,02 (mol)
n HCl =0.15.0.2 = 0.03 (mol) •
Phifdng trinh dien l i cdc axit:
H2SO4 — ^ 2H* + SỖ • 0.01 mol 0,02 mol . . 0.01 mol 0,02 mol . . HNO3 > H^ + NO3 . 0,02 mal 0,02 mol HCl — > H^ + cr . 0,03 mol 0,03 mol => I = 0.02 + 0.02 + 0.03 = 0.07 (mol) Goi the tich dung djch B can diing la V.
HNaOH = 0,3.V (mol)
nKOH = 0.l'5.V(mol)
NaOH > Na* + OH" 0.3V (mol) 0,3V (mol) KOH > K* + OH"
0,15V (mol) 0,15V (mol) => Z n _ . = 0,3V + 0,15V = 0,45 V (mol)
OH
Phan tfng trung hoa dung dich A va dung dich B : H* + O H - > H2O .
. 0,45V 0.45V (mol)
Dung dich thu dufdc cd pH = 3 => moi trtfcfng axit ^^H"" di/.
The tich dung dich sau phan tfng : V j j = 0,6 + V pH = 3 [ H i = 10-'= 0.001 (M)
[HI = ^ = "'Q^-Q''^^^ = 0.001 V = 0.154 lit V 0.6 + V
Vay the tich dung dich B la 0,154 lit.
Bai 9. Trpn 200ml dung dich g6m HCl ỌIM va H2SO4 0.05M vdi 300ml dung dich Bă0H)2 cd nong dp a mol// thu di/dc m gam ket tua va 500ml dung dich I cd pH = 13. Tinh a va m. Cho biet trong cac dung dich vdi dung moi la niTdc,
tichso [ H l . [ O H - ] = 1 0 - ^
Phan dgng va phuong ph^p giii H6a hpc 11 VP CO - 05 Xuan Hi/ng
Gidi
Ta CO: nHci = 0,1.0.2 = 0,02 (mol); nH2S04 -0,05.0,2 = 0,01 (mol) HCl — > H^ + cr HCl — > H^ + cr
0,02 mol 0,02 mol
H2SO4 > 2H^ + SO4"
0,01 mol 0,02 mol 0,01 mol =>In + =0,02+ 0,02 = 0,04 (mol) =>In + =0,02+ 0,02 = 0,04 (mol) Vataco nB^(OH)2 =0'3.ămol) Bă0H)2 > Bâ* + 20H- 0,3a mol 0,3a mol 0,6a mol
Khi trpn dung dich gom HCl va H2SO4 vdi dung dich Bă0H)2 ta c6 cac phúdng trinh sau : phúdng trinh sau :
+ OH" > H2O (1) 0,04 0,04 0,04 0,04
Bâ* + SÔ- > BaS04i (2) 0,01 0,01 0,01 0,01 0,01 0,01
Dung dich CO pH= 13 [Hi = lỐ^M
=>[0H-]= - i — = 10"' =0,1M
iõ'-^
=> n _ =0,1.0,5 = 0,05 (mol)
OH M
Tac6: n _ = 0,6a-0,04 OH dir
o 0,6a - 0,04 = 0,05 o a = 0,15 (M) n„ 2+ = 0,3a = 0,3.0,15 = 0.045 (mol) n„ 2+ = 0,3a = 0,3.0,15 = 0.045 (mol) Bu Ttr(2)=:>n 2+ = 0.045-0.01 = 0,035(mol) => ng^sọ =0.01 mol => m = mB„s04 = 233.0,01 = 2.33 (gam). B^i 10.
a) Dung dich CH3COOH ỌIM c6 dp di^n li a = 1%. Vifít phUdng trinh dien li
CHiCOOH va tinh pH cua dung dich nhỵ
b) A Ih dung djch HCl 0.2M; B la dung dich H2SO4 ỌIM. Trpn cac the tlch b<^ng nhau cua A va B diTpc dung dich X. Tinh pH cua dung dich X. Cho lg4 = 0,6; nhau cua A va B diTpc dung dich X. Tinh pH cua dung dich X. Cho lg4 = 0,6;
lg2 = 0.3. (DHQG.d0l)
Gidi
a) PhiTdng trinh dien li: CH3COOH - » CH3COO- +
BandSu: 0,1 0 0 Can bang: 0 . 1 - x x x Can bang: 0 . 1 - x x x -.100% = 1% =^ X = 10-' hay [H^ = IQ-' (M) (M) (M) a = 0,1 =>pH = -lg[Hl = -lgl0-' = 3.
b) The tich dung djch sau khi trpn : Vjj = VA + VB = 2.V (lit)
OHci = 0,2.V (mol) HCl — > H^ + cr 0,2V 0,2V nH2S04 =0,l.V(mol) H2SO4 > 2H* + SÔ" 0,1.V 0,2.V =>In„. = 0,2V + 0,2V = 0,4V (mol) H [Hi = V 2V 0,4V = 0,2(M) Vay pH = -lg[Hl = -lgO,2 = 0,7.
Bai 11. Trpn dung dich X chuTa NaOH 0,1M, Bă0H)2 0,2M vdi dung dich Y (HCl 0,2M; H2SO4 0,1M) theo ti Ip nao ve the tich de dung dich ihu diTpc c6 (HCl 0,2M; H2SO4 0,1M) theo ti Ip nao ve the tich de dung dich ihu diTpc c6 pH= 13?
Gidi
Taco: nB,(OH)2 = 0,2Vx molj
" N a O H = 0,lVx mol J ~ " ^ ' 0 H -
nH2S04 = 0,1VY mol T
"HCI - 0,2VY mol
Sau khi trpn dung dich c6 pH = 13 moi triTclng bazd => sau phan ifng giiTa axit va bazd thi OH" diT: axit va bazd thi OH" diT:
H^ + OH" -> H2O 0,4VY 0,4VY 0,4VY 0,4VY = 0,4VY mol n^ , ^ . u . = (0,5VX-0,4VY) mol Tac6:pH= 13 => [ H i = 1 0 1-13 [0H1 = 10 ,-1 0,5VY -0,4VX VX _ 4 Vx + VY 91
Phfln-djino va phuong ph^p gSi HdaJipc 11 VP co - D5 Xuan Hifng
Hkil2. Tron iOOmI dung^ljch g6m BăOH)2 ỌIM va NaOH 0,1M vdi 400 ml dung dich gom H2SO4 0/)375M va HCl 0,0125M thu di/dc dung djch X. Gia tri pH cua dung dlchX 1^:
Ạ 2 B. 1 C . 6 D. 7.
(Trich de thi tuyen sink Dai hoc khoi B) Gidi
Ta c6: nBăOH)2 = ÓÔ mol
^ = > > I . O H - "NaOH = 0,01 mol nH2S04 = 0.015 mol -j " H C I = 0,005 mol Zn = 0,03 mol Tn. . = 0,035 mol
Khi tron Ian hon hdp 2 axit 2 bazd xay ra phan xing trung hoa: H* + OH" H2O 0,03 0.03 n^^. = 0.035 - 0,03 = 0,005 mol H •HI 0.005 0,5 = 0,01 = 10-2 => pH = 2 => Dap an Ạ
B a i 13. Cho m gam hon hdp Mg va Al vio 250ml dung djch X chiJa hon help axit HCl I M axit H2SO4 0.5M thu difdc 5,32 lit Hj (dktc) va dung dich Y (coi the tich dung dich khong doi). Dung dich Y c6 pH la:
Ạ 7 B. 1 C. 2 D. 6.
(Trich de tlii tuyen sinh Bai hoc khoi A)
cm Ta c6: HHCI = 0,25 mol Ta c6: HHCI = 0,25 mol " H 2 S 0 4 = 0,125 mol 5,32 nH2 = 22,4 = 0,2375 mol mol
Khi cho Mg, Al tac dung vdi hon hdp 2 axit HCl va H2SO4, ta c6 sd do phan «?ng: 2H* -> H2 0,475 , 0,2375 => = 0,5 - 0,475 = 0,025 mol n + -1 0,025 H^ 0,25 Dap an B . = 0.1 = 10"' M pH = 1 KifAwrfiinET D a n a 3.