Find the ph of the following solution: 6.00 ml of 3.00 m hno3 diluted to a volume of 18.0 liters with water

The first thing to do is to calculate the new concentration of the solution after dilution. We do as follows:

M1V1 = M2V2
(3.0 M).006 L  = M2 (18 L)
M2 = 0.001 M HNO3

When in solution it dissociated into ions as follows:

HNO3 = H+ + NO3-

[H+] = 0.001 M HNO3 ( 1 mol H+ / 1 mol HNO3 ) = 0.001 M H+

pH = -log[H+] = -log 0.001 = 3

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