Calculate d pH of 0.005mol/dm³ tetraoxosulphate(vi)
To calculate the pH of a tetraoxosulphate(VI) solution, you first need to determine the concentration of the hydrogen ion.
The chemical formula for tetraoxosulphate(VI) is H2SO4. When tetraoxosulphate(VI) dissociates in water, it produces two hydrogen ions (H+) and one sulfate ion (SO42-).
H2SO4 → 2H+ + SO42-
The concentration of hydrogen ions in a 0.005 mol/dm³ tetraoxosulphate(VI) solution is therefore:
[H+] = 2 x 0.005 mol/dm³
[H+] = 0.01 mol/dm³
To calculate the pH of the solution, you can use the following formula:
pH = -log[H+]
pH = -log(0.01)
pH = 2
Therefore, the pH of a 0.005 mol/dm³ tetraoxosulphate(VI) solution is 2.
To calculate the pH of a tetraoxosulfate (VI) solution, we need to consider the dissociation of the tetraoxosulfate (VI) ion, also known as sulfate (SO4^2-), in water.
The formula for the dissociation of sulfate is as follows:
SO4^2- + H2O ⇌ HSO4^- + OH^-
Since tetraoxosulfate (VI) is a strong base, it dissociates completely in water to form hydroxide ions (OH^-). Therefore, the concentration of hydroxide ions in the solution is equal to the concentration of tetraoxosulfate (VI).
Given that the concentration of tetraoxosulfate (VI) is 0.005 mol/dm^3, the concentration of hydroxide ions is also 0.005 mol/dm^3.
To calculate the pOH of the solution, we can use the logarithmic definition:
pOH = - log10[OH^-]
Substituting the value for the hydroxide ion concentration:
pOH = - log10(0.005)
Using a calculator, we can find the pOH to be approximately 2.301.
Since pH + pOH = 14 (at room temperature and for dilute solutions), we can calculate the pH by subtracting the pOH from 14:
pH = 14 - 2.301
Therefore, the pH of the 0.005 mol/dm^3 tetraoxosulfate (VI) solution is approximately 11.699.