Question

calculate the [H⁺] and the pH of each of the solutions. {Show calculation}
a) 0.314 M HNO₃    pH =          :[H⁺] =
d) 0.314 M C₂H₅NH₂    pH =          :[H⁺] =
e) 0.314 M HF    pH =          :[H⁺] =
j) 0.314 M Ba(OH)₂    pH =          :[H⁺] =

Answer 1

HNO3 is strong acid so it can completely dissociated into its ions, so [H+] = 0.314

pH = -log (0.314)

pH = 0.5031

C2H5NH2 ia weak base and its pKb value = 3.19

pOH = 1/2 (pKb -log C)

pOH = 1/2 ( 3.19 - log(0.314))

pOH = 1.846

pH = 14 - 1.846 = 12.154

[H+] = 10^-12.154 = 7.014 * 10^-13

HF is a wean acid and its pKa = 3.17

pH = 1/2 (pKa - log C)

pH = 1/2 ( 3.17 - log (0.314))

pH = 1.836

[H+] = 10^-1.836 = 0.01459

Ba(OH)2 is strong base and it can dissociate into its ions completely,

pOH = - log [OH-]

pOH = -log (0.314)

pOH = 0.5031

pH = 14 - 0.5031

pH = 13.4969

[H+] = 10^-13.4949 = 3.2 * 10^-14