1
a.
stim ca [H3O+]*[HO-] = 10⁻¹⁴
=> [HO-] = 10⁻¹⁴/[H3O+]
= 10⁻¹⁴/2*10⁻⁸ = 5*10⁻⁷ mol/L
caracter bazic, deoarece [H3O+] < [HO-]
b.
stim ca [H3O+]*[HO-] = 10⁻¹⁴
=> [HO-] = 10⁻¹⁴/[H3O+]
= 10⁻¹⁴/10⁻¹³ = 10⁻¹ mol/L
caracter bazic, deoarece [H3O+] < [HO-]
c.
stim ca [H3O+]*[HO-] = 10⁻¹⁴
=> [HO-] = 10⁻¹⁴/[H3O+]
= 10⁻¹⁴/4*10⁻⁴ = 2,5*10⁻¹¹ mol/L
caracter acid, deoarece [H3O+] > [HO-]
2.
a.
stim ca [H3O+]*[HO-] = 10⁻¹⁴
=> [H3O+] = 10⁻¹⁴/[HO-]
= 10⁻¹⁴/4*10⁻³ = 2,5*10⁻¹² mol/L
caracter bazic, deoarece [H3O+] < [HO-]
b.
stim ca [H3O+]*[HO-] = 10⁻¹⁴
=> [H3O+] = 10⁻¹⁴/[HO-]
= 10⁻¹⁴/10⁻⁹ = 10⁻⁵ mol/L
caracter acid, deoarece [H3O+] > [HO-]
c.
stim ca [H3O+]*[HO-] = 10⁻¹⁴
=> [H3O+] = 10⁻¹⁴/[HO-]
= 10⁻¹⁴/10⁻³ = 10⁻¹¹ mol/L
caracter bazic, deoarece [H3O+] < [HO-]
