ph (1 Viewer)

[9876543210]

I need help, please someone....

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A 2.3 × 10–4 mol/L solution of sulfuric acid is completely dissociated. Calculate its pH.
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b​

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The pH of an 0.020 mol/L solution of sulfuric acid was 1.58. What can you deduce about the extent of ionisation of the HSO4
– ion in this solution?

answers:
a) 3.34
b) 32% ionised

thx!​

 

[9876543210]

please someone help me...
i have an exam tomorrow!!!
i know that im sounding really desperate

 

[clintmyster]

I need help, please someone....

a​

A 2.3 × 10–4 mol/L solution of sulfuric acid is completely dissociated. Calculate its pH.
​

b​

The pH of an 0.020 mol/L solution of sulfuric acid was 1.58. What can you deduce about the extent of ionisation of the HSO4
– ion in this solution?

answers:
a) 3.34
b) 32% ionised

thx!​

a) ph = -log₁₀[2 x 2.3 × 10^–4] = 3.34

b) Not sure if this is the right way to do it but i get a different answer anyways. Hopefully someone can correct my mistake.

H⁺ = 10^-ph
= 10^-1.58
= 0.02630267992 M
Therefore H⁺ concentration for Sulfuric acid is 2x 0.0263...= 0.0526053598M

Therefore degree of ionisation is (0.02 / 0.0526) x 100 = 38% ionised

 

[annabackwards]

a) ph = -log₁₀[2 x 2.3 × 10^–4] = 3.34

b) Not sure if this is the right way to do it but i get a different answer anyways. Hopefully someone can correct my mistake.

H⁺ = 10^-ph
= 10^-1.58
= 0.02630267992 M
Therefore H⁺ concentration for Sulfuric acid is 2x 0.0263...= 0.0526053598M

Therefore degree of ionisation is (0.02 / 0.0526) x 100 = 38% ionised

I would do it like you did.

 

[clintmyster]

I would do it like you did.

hopefully he made a typo then haha

 

[jet]

Heres how i would do b)
The pH of the soln is 1.58
Therefore [H⁺] = 10^-1.58
=0.0263

Now, assuming that the first H⁺ ions were removed from the H₂SO₄ molecule (as it is a strong acid), then there should be 0.02 M of these in solution.
Hence, the number of H⁺ ions released in solution from the HSO₄ ion is 0.0263 - 0.02 = 0.063 of the 0.02 ions there are

Therefore % ionisation of the HSO₄ ion= [0.0063/0.02] x 100 = 31.5 = 32%

 

[clintmyster]

Heres how i would do b)
The pH of the soln is 1.58
Therefore [H⁺] = 10^-1.58
=0.0263

Now, assuming that the first H⁺ ions were removed from the H₂SO₄ molecule (as it is a strong acid), then there should be 0.02 M of these in solution.
Hence, the number of H⁺ ions released in solution from the HSO₄ ion is 0.0263 - 0.02 = 0.063 of the 0.02 ions there are

Therefore % ionisation of the HSO₄ ion= [0.0063/0.02] x 100 = 31.5 = 32%

Oh whoa daym didn't think of that. I did something very bad and did not right my daym eqn. My teacher told us that it would haunt us one day and it now it happened haha. Good Job.

 

[jet]

Thanks

 

[9876543210]

oo....
thx!
it helped!
=]

 

[4theHSC]

I need help, please someone....

a​

A 2.3 × 10–4 mol/L solution of sulfuric acid is completely dissociated. Calculate its pH.
​

b​

The pH of an 0.020 mol/L solution of sulfuric acid was 1.58. What can you deduce about the extent of ionisation of the HSO4
– ion in this solution?

answers:
a) 3.34
b) 32% ionised

thx!​

LOL i went through that question... conquering chem... the idea is that there are 2 equations, one completely dissociates and the other one doesn't... I heard that in the syllabus they usually assume that both these reactions go to completion...