Solution :
A) 2.3 M HNO3
HNO3 is dissociated as,
HNO3 === H+ + NO3-
2.3 M == 2.3 M --- 2.3 M
Thus, [H+] = 2.3 M
pH = - log [H+] = - log 2.3
pH = - 0.36
B) 0.72 M Ba(OH2
Ba(OH)2 === Ba2+ + 2OH-
0.72 M ==== 0.72 M --- 2 x 0.72 M
Thus,
[OH-] = 2 x 0.72 = 1.44
pOH = - log [OH-] = - log 1.44 = - 0.16
Since,
pH + pOH = 14
pH = 14 - pOH
pH = 14 - (-0.16) = 14 + 0.16 = 14.16
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