Odpowiedź:
Wyjaśnienie:
Dane: Szukane:
C KOH = 0,001mol/dm³ pH, pOH, [H⁺]=?
KOH to mocny elektrolit, czyli C=[OH⁻]
[OH⁻] = 10⁻³mol/dm³
Korzystam ze wzoru na iloczyn jonowy wody:
[H⁺]*[OH]⁻ = 10⁻¹⁴
[H⁺] = 10⁻¹⁴ : 10⁻³
[H⁺] = 10⁻¹¹mol/dm³
pH = -log[H⁺]
pH = -log(10⁻¹¹)
pH = 11
pH + pOH = 14
pOH = 14 - 11
pOH = 3
II sposób:
pOH = -log[OH⁻]
pOH = -log(10⁻³)
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Odpowiedź:
Wyjaśnienie:
Dane: Szukane:
C KOH = 0,001mol/dm³ pH, pOH, [H⁺]=?
KOH to mocny elektrolit, czyli C=[OH⁻]
[OH⁻] = 10⁻³mol/dm³
Korzystam ze wzoru na iloczyn jonowy wody:
[H⁺]*[OH]⁻ = 10⁻¹⁴
[H⁺] = 10⁻¹⁴ : 10⁻³
[H⁺] = 10⁻¹¹mol/dm³
pH = -log[H⁺]
pH = -log(10⁻¹¹)
pH = 11
pH + pOH = 14
pOH = 14 - 11
pOH = 3
II sposób:
pOH = -log[OH⁻]
pOH = -log(10⁻³)
pOH = 3