Question

il vreau pe 5.7.8.10

Answer 1

10. pH = ?

V = 0,224 L HCl
V H2O = 100mL 

C =V/Vm = 0,224/22,4 = 0,01 moli HCl

C = n/Vs = 0,01/0,1 = 0,1 moli/L

[HCl] = C = [H3O+] = 10^(-pH) = 10^(-1) => pH = 1

8. Vs1 = ? Ca(OH)2
C1 = 0,2M
Vs2 = 500mL = 0,5 L HCl
C2 = 0,3M
pH final = 7

C2 = n2/Vs2 => n2 = C2*Vs2 = 0,3*0,5 = 0,15 moli HCl

1 mol..........2 moli
Ca(OH)2 + 2HCl => CaCl2 + 2H2O
x moli........0,15 moli

x = 0,15/2 = 0,075 moli Ca(OH)2

C1 = n1/Vs1 => Vs1 = n1/C1 = 0,075/0,2 = 0,375 L = 375mL sol. Ca(OH)2

7. Vs1 = 1V KOH
C1 = 3×10⁻⁴ M
Vs2 = 9,7V
C2 = 10⁻³ M

C1 = n1/Vs1 => n1 = C1*Vs1 = 3×10⁻⁴×1V
C2 = n2/Vs2 => n2 = C2*Vs2 = 10⁻³×9,7V

V final = 9,7V+1V = 10,7V
n final = n1+n2 = 3×10⁻⁴×1V+10⁻³×9,7V

C final = n final/V final = (3×10⁻⁴×1V+10⁻³×9,7V)/(10,7V) = 13×10⁻⁴ M

[OH-] = C = [KOH] =13×10⁻⁴ M => pOH= -lg 13*10^(-4) = 2,886

pH = 14-2866 = 11,113

5. a)
ms = ro*V = 1,15V
md = C*ms/100 = 3*1,15V/100 = 0,0345V

C = md/M*V = 0,0345V/36,5V= 0,0009452 M 

HCl -> A.T => [HCl] = C = [H3O+] = 0,0009452

pH = -log [H3O+] = -lg 0,0009452 = 3,0244

pH + pOH = 14 => pOH = 10,975

b) V presupus = 1L = 1000cm3

ro = 1,02*1000 = 1020g/L
md = ms*C/100 

 C = md/M*Vs = ms*C/M*Vs*100 = C*ro/M*100 = (0,0392*1020)/(40*100) = 0,01 M

NaOH -> B.T => [NaOH] = C = [OH-] = 10⁻pOH = 10⁻² M ⇒ pOH = 2 

pH + pOH = 14 ⇒ pH = 12