Table of Contents
What is the pH of 0.05 N H2SO4?
Therefore the pH value of a 0.005 molar aqueous solution of sulphuric acid is approximately 2.0.
What is the pH of 0.005 m H2SO4?
Hence, pH is 2.
What is the pH of 0.1 N H2SO4?
And so 0.1⋅mol⋅L−1 is 0.2⋅mol⋅L−1 with respect to H3O+ . And thus pH=−log10(0.2)=−(−0.699)=0.699 . And on the other hand, 0.1⋅N sulfuric acid is in fact 0.05⋅mol⋅L−1 with respect to sulfuric acid….the acid STILL requires TWO equivs of base for stoichiometric equivalence.
How do you make 0.05 N H2SO4?
- Take 950 ml distilled and cool water in a container.
- Take exactly 2.45 ml( for 0.05 N H2SO4) or 4.9 ml ( for 0.05 M H2SO4 ) pure Sulphuric acid (H2SO4) .
- Now, add Sulphuric acid to the distilled water drop by drop with constant stirring.
- Allow the solution to be cooled.
- Add some more water to make it upto 1000 ml.
What is pH of 1n H2SO4?
Specifications
Color | Undesignated |
---|---|
pH | 1.0 |
Molecular Formula | H2O4S |
UN Number | 2796 |
InChI Key | QAOWNCQODCNURD-UHFFFAOYSA-N |
What is the pH of a 5m H2SO4?
OK, so it probably is a homework question (the what is the pH of a 5M H2SO4) and yes, using the standard definition for pH (pH = -log [H+]) a diprotic acid, fully dissociated (e.g., H2SO4) at a concentration of 5 M will give 10 M H+ concentration (though perhaps a different activity) and leads to a negative pH (= -1 here).
What is the concentration of sulfuric acid in H3O+ at pH 1?
Therefore, concentration of sulfuric acid should be a half of concentration of H 3 O +. concentration of sulfuric acid at pH 1 is 0.05 mol dm -3 Sulfuric acid shows very low pH value.
What is the concentration of 1H in H2SO4?
1 H concentration = H 2 SO 4 concentration 2 H + concentration = 0.00896 M More
Is H2SO4 a strong acid or a weak acid?
H2SO4 is a strong polyprotic acid. It undergoes 2 deprotonations, only one of which is complete. They are as follows: is a strong acid. The molar concentration of H3O+ and HSO4- are thus both equivalent to the analytical (initial) concentration of H2SO4, which, in this case is 0.1 mol/L.