Calculate the pH for .455 M HNO3?

2007-03-16 02:21:50 · 2 answers · asked by lilballerina2004 1 in Science & Mathematics ➔ Chemistry

pH = - log[H+]
pH = - log[0.455]

2007-03-16 02:29:43 · answer #1 · answered by tuoidabuon 2 · 0⤊ 0⤋

HNO3--->H+ +NO3- strong acid completely dissociated

So [H+]= 0.453

pH = log 1/[H+] = log1/0.455=log 2.20 =0.342

2007-03-16 02:47:04 · answer #2 · answered by maussy 7 · 0⤊ 0⤋

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