- log(0.000626 M H2SO4)
= 3.2 pH
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Assuming pH on just simple concentration. - log(0.01 M H2SO4) = 2 pH =====
30 mL water.
You need 49,8 mL H2SO4 6,4M.
The first solution is more concentrated because it contains 6 moles of H2SO4 per one liter of solution. The second solution is less concentrated because it contains 0.1 moles of H2SO4 in one liter. In equal amounts of each example, the first would have more H2SO4.
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
Assuming pH on just simple concentration. - log(0.01 M H2SO4) = 2 pH =====
30 mL water.
You need 49,8 mL H2SO4 6,4M.
The first solution is more concentrated because it contains 6 moles of H2SO4 per one liter of solution. The second solution is less concentrated because it contains 0.1 moles of H2SO4 in one liter. In equal amounts of each example, the first would have more H2SO4.
pH = - log[H+] so a 0.01 M solution of HCl has, pH= 2
its PH is 3
0.08 n
Molarity = moles of solute/volume of solution 0.324 M H2SO4 = moles H2SO4/500 ml 162 millimoles, or, more precisely to the question 0.162 moles H2SO4
-log(0.5 M HF) = 0.3 pH
- log(0.12 M) = 0.92 pH ---------------
A 0.01 M solution of NaOH has a pH =13
Its pH value is 2.