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. (adsbygoogle = window.adsbygoogle || []).push({});
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; (adsbygoogle = window.adsbygoogle || []).push({});
- 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) (adsbygoogle = window.adsbygoogle || []).push({});
[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? (adsbygoogle = window.adsbygoogle || []).push({});
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.