1,obliczyc poh roztworu HNO3 o stężeniu 2,3*10^-4 mol/dm^3
2,obliczyc ph roztworu HCl o stężeniu 1,3*10^-4 mol/dm^3
HNO₃<->H⁺+NO₃⁻Zatem na 1 mol HNO₃ przypada 1 mol H⁺Czyli [H⁺]=2,3*10^-4pH=-log[H⁺]=-log[2,3*10^-4]=3,64Zatem pOH=14-3,64=10,36HCl->H⁺+Cl⁻1mol HCl-1mol H⁺[H⁺]=1,3*10^-4pH=-log[H⁺]=-log[1,3*10^-4]=3,88
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HNO₃<->H⁺+NO₃⁻
Zatem na 1 mol HNO₃ przypada 1 mol H⁺
Czyli [H⁺]=2,3*10^-4
pH=-log[H⁺]=-log[2,3*10^-4]=3,64
Zatem pOH=14-3,64=10,36
HCl->H⁺+Cl⁻
1mol HCl-1mol H⁺
[H⁺]=1,3*10^-4
pH=-log[H⁺]=-log[1,3*10^-4]=3,88