Determine the pH of each of the following solutions?

0.025M HI --> 0.025M H+.....-log(0.025) = pH = 1.6

0.116M HClO4 --> 0.116M H+.....-log(0.116) = pH = 0.94

solution of HClO4 + HCl will have an additive amount of H+

assuming 100ml for both acids since volume is not stated

0.0054moles H+ from HClO4 + 0.003moles H+ from HCl = 0.0084moles H+

total volume = 200ml

0.0084moles / 0.2L = 0.042M H+.....-log(0.042) = pH = 1.38

1.8% HCl

100ml of 1.8% HCl = 101g.....100ml x 1.01g/ml

1.8% of 101g = 1.818g HCl

1.818g / 36.45g/mole = 0.05moles HCl.....0.05moles H+ / 0.1L = 0.5M.....pH = 0.3

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Determine the pH of each of the following solutions?

Determine the pH of each of the following solutions:

2.5x10^-2 M HI

0.116 M HClO4

a solution that is 5.4×10−2 M in HClO4 and 3.0×10−2 M in HCl

and a solution that is 1.80% HCl by mass (Assume a density of 1.01 g/mL for the solution).

Please give an explanation with answers! I really...

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add the two concentrations together (4.4x10^-2) etc, and that will give you the total concentration of H+ ions. ( 4.4×10^−2)+(5.6×10^−2) = ??? Now to find the pH take the -log [H+] so.... pH=-log[( 4.4×10^−2)+(5.6×10^−2)] =??? for the second one.. 1.08% by mass = 1.08 g HCLO4/100g solution. Multiply by different conversion factors to get moles/liters (not mL). This will give you [H+] so again... pH=-log[H+]