If pH=3.31, then find out [H'] (Approxy)(1) 3.39*10-4(3) 3.0 x 103(2) – InterviewSolution

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1.

If pH=3.31, then find out [H'] (Approxy)(1) 3.39*10-4(3) 3.0 x 103(2) 5 x 104(4)

Answer»

pH =-log[H+]

=> -3.31 = log([H+])=> log[H+] = -4+0.69=> [H+] = 10^-4×10^0.69 = 5×10^-4

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