Question

am folosit regula triunghiului, mi a dat 0.25parti=Vs sol cu cM=0.25 si 0.15parti=Vs sol cu cM=0.35
asta inseamna ca 0.25p+0.35=2000ml
p=3333.33ml
este corect pana aici?

(pH=12 deci e solutie bazica
asa ca pOH=2, cM=10 la-2=0.1M

am reactia
KOH+HNO3<->KNO3+H20 -neutralizare
trebuie sa scriu mai scriu cumva o reactie?
KOH->K+ + OH-

Answer 1

pH =  12 
Vsf = 2L

pH+pOH = 14 ⇒ pOH = 2

[OH⁻] =$10^{-pOH}$ = 10⁻² M = 0,01M

C₁ = 0,25M...........................Vs₁ = 0,34 parti volum
                     Cf = 0,01M
C₂ = 0,35M...........................Vs₂ = 0,24 parti volum..

parti V total = 0,58

0,58 parti V................0,34...................0,24 parti
2L....................Vs₁ = 1,172L KOH............Vs₂ = 0,827L HNO₃


C₁ = n₁/Vs₁ ⇒ n₁ = C₁×Vs₁ = 0,25Vs₁ = 0,25*1,172 = 0,293 moli KOH
C₂ = n₂/Vs₂ ⇒ n₂ = C₂×Vs₂ = 0,35Vs₂ = 0,35*0,827 = 0,28945 moli HNO3

m KOH = 0,28945*56 = 16,2g

m KNO₃ = 101*0,28945  = 29,25g

Cf = (n₁+n₂)/(Vs₁+Vs₂) ⇔ 0,01 =[(0,25Vs₁)+(0,35Vs₂)]/(Vs₁+Vs₂) ⇔ 0,01Vs₁+0,01Vs₂ = 0,25Vs₁+0,35Vs₂ ⇔ 0,24Vs₁ = 0,34Vs₂ ⇒ Vs₁/Vs₂ = 34/24 = 17/12 ⇒ Vs₂ = 12Vs₁/17 = 0,705Vs₁

Vs₁+Vs₂ = 2 ⇔ 1,705Vs₁ = 2 ⇒ Vs₁= 1,172 L sol. KOH

Vs₂ = 0,827 L sol. HNO₃

C₁ = n₁/Vs₁ ⇒ n₁ = C₁*Vs₁ = 0,25*1,172 = 0,293 moli KOH

m KOH = 56*0,293 = 16,4g 

C₂ = n₂/Vs₂ ⇒ n₂ = C₂*Vs₂ = 0,35*0,827 = 0,2896 moli HNO₃

KOH + HNO₃ ⇒ KNO₃ + H₂O

n HNO₃ = n KNO₃ = 0,2896 moli

m KNO₃ = 0,2896*101 = 29,25g

A,B,C - adevarat  , D, E - fals