Obliczyć pH roztworów
a) Ba(OH)2 o stężeniu 3*10^-5M
b) KOH o stęz. 0.02M
c)HCl o stez 0.02M
d)H2SO4 o stez 0.05M
e)CH3COOH o stęz. 0.01M, K=1,75x10^-5
f)NH3 x H2O o stez. 0.01M a=2%
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Zadanie 1.
a)
Ba(OH)₂ ⇒ Ba²⁺ + 2OH⁻
_______ _____
1mol : 2mol
3*10⁻⁵M : 6*10⁻⁵M
pOH=-log[OH⁻]
pH + pOH = 14
pH = 14 - pOH
pH = 14 - (-log[OH⁻])
pH=14 + log[OH⁻]
pH = 14 + log6*10⁻⁵ = 14 + ( 0,78 - 5 ) = 14 - 4,22 = 9,78
Odp.: pH 0,00003M r-r Ba(OH)₂ wynosi 9,78.
b)
KOH ⇒ K⁺ + OH⁻
____ ____
1mol : 1mol
0,02M : 0,02M
pH = 14 + log[OH⁻]
pH = 14 + log0,02 = 14 -1,7 = 12,3
Odp.: pH = 12,3
c)
HCl ⇒ H⁺ + Cl⁻
__ ___
1mol:1mol
0,02M:0,02M
pH=-log[H⁺]
pH=-log0,02=-(-1,7) = 1,7
Odp.: pH = 1,7
d)
H₂SO₄ ⇒ 2H⁺ + SO₄²⁻
_____ ____
1mol : 2mol
0,05M : 0,1M
pH = -log[H⁺]
pH = -log0,1 = - (-1 ) = 1
Odp.: pH=1
e)
CH₃COOH ⇔ CH₃COO⁻ + H⁺
[CH₃COO⁻]=[H⁺]=x