Calculate the pH of 0.005moldm-3 tetraoxosulphate(VI)
I assume you mean H2SO4. The correct IUPAC name is sulfuric acid.
Is this a beginning or an advanced class?
Beginning:
...............H2SO4 --> 2H^+ + SO4^2-
(H^+) = 2*0.005 = 0.01
pH = -log 0.01 = 2
Advanced:
H2SO4 is a diprotic acid. The first H^+ is a strong acid but the second H^+ is weak.
..............H2SO4 ==> H^+ + HSO4^- complete ionization so H^+ = 0.005 M
..............0.005.......0.005.....0.005
Then HSO4^- ==> H^+ + SO4^2- for which k2 = 0.012
I.........0.005...........0.005.............0
C..........-x...............x..............x
E........0.005-x.....0.005+x.......x
You need to use the k2 in you text/notes. My text says 0.012
k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
Substitute (0.005+x)(x)/(0.005-x). Solve the quadratic and final (H^+) = 0.005 + x. Then convert to pH. I obtained 0.003 for the second ionization so total H^+ = 0.005+0.003 = 0.008 for pH = 2.1
Compare the two sets; i.e., beginning H^+ = 0.01 and pH 2 with advanced H^+ = 0.008 and pH 2.1.
I don't understand
I am lost
Calculate the ph of h2so4 with 0.005mol/dm3
Calculate the PH of 0.01M tetraoxosulphate vi acid solution
The correct formula for tetraoxosulphate(VI) acid is H2SO4.
To calculate the pH of a 0.01M H2SO4 solution:
H2SO4 dissociates in water as follows:
H2SO4 → 2H+ + SO4^2-
The concentration of H+ ions in the solution is 2 x 0.01 = 0.02 M.
pH = -log[H+] = -log(0.02) = 1.7
Therefore, the pH of a 0.01M H2SO4 solution is 1.7.
To calculate the pH of a solution of tetraoxosulphate(VI) (also known as sulfuric acid, H2SO4), we need to know the concentration of the acid. However, you have provided the concentration in moles per cubic decimeter (mol/dm³), so we can use this value directly.
The pH of an acidic solution can be calculated using the formula:
pH = -log[H+]
where [H+] represents the concentration of hydrogen ions in the solution.
In the case of tetraoxosulphate(VI), it is a strong acid that completely dissociates in water. This means that for every mole of H2SO4, there will be two moles of H+ ions produced.
Given that the concentration of H2SO4 is 0.005 mol/dm³, the concentration of H+ ions is twice that value:
[H+] = 2 * 0.005 = 0.01 mol/dm³
Now, we can plug this concentration into the pH formula:
pH = -log(0.01) ≈ 2
Therefore, the pH of the 0.005 mol/dm³ tetraoxosulphate(VI) solution is approximately 2.