Pv =nRT
T= 273
n = 0.500
v= 11.2
R= 0.08206
p= 0.5×0.08206×273 ÷ (11.2) =10.00
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?
n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)
c₂v₂=2c₁v₁
c₂=2c₁v₁/v₂
c₂=2*2.00*0.25/2.00=0.5 mol/L
0.5 M NaOH
Bromcresol green is the indicator that is blue in a solution that has a Ph of 5.6.
Ksp of AgCl= 1.6×10⁻¹⁰
AgCl=Ag⁺ +Cl⁻
Ksp=[Ag⁺][Cl⁻]
Assume [Ag⁺]=[Cl⁻]=x
Ksp=x²
1.6×10⁻¹⁰=x²
x=0.000012
In FeCl₃:
FeCl₃------>Fe⁺³+ 3Cl⁻
as there is 0.010 M FeCl₃
So there will be ,
[Cl⁻]= 0.030
So
[Ag⁺]=Ksp/[Cl⁻]
=1.6×10⁻¹⁰/0.030
=5.3×10⁻⁹
so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.
