In: Chemistry
Calculate [H3O+] in the following solutions.
a)3.0×10−2 M HCl and 7.0×10−2 M HOCl
b)0.105 M NaNO2 and 5.50×10−2 M HNO2
c)5.30×10−2 M HCl and 7.64×10−2 M NaC2H3O2
a)
concentration of [H3O+] = 3.0 x 10^-2 M
b)
pH = pKa + log [NaNO2 / HNO2]
= 3.25 + log [0.105 / 5.50×10−2]
= 3.53
[H3O+] = 10^-3.53
[H3O+] = 3.0 x 10^-4 M
3) 5.35×10−2 M HCl and 7.66×10−2 M NaC2H3O2
C2H3O2- + HCl ---------------> CH3COOH + Cl -
7.64×10−2 5.30×10−2 0
2.34 x 10^-2 0 5.30×10−2
pH = pKa + log [2.34 x 10^-2 / 5.30×10−2]
= 4.74 + log [2.34 x 10^-2 / 5.30×10−2]
= 4.38
[H3O+] = 4.2 x 10^-5 M