Calculate the hydrogen ion concentration in this solution?

A 0.21g sample of sulphuric acid is dissolved completely in sufficient water to make 0.25 litre of final solution. I have to calculate the hydrogen ion concentration and show steps of working it out and then say what the pH of the sulphuric acid solution is.

Answer:
The molar mass of sulphuric acid is 98.
If there are 0·21g of the acid in 0·25 litre of solution, there will be 4 times this amount in a litre.
4 x 0·21 = 0·84 g per litre
Hence the molarity is 0·84 ÷ 98 mol per litre
This comes to 0·00571 mol per litre.
But the acid is diprotic.
Assuming it to be 100% ionised (a reasonable assumption at this dilution), [H+] = 2 x 0·00571 = 0·01714 mol/litre.
I haven't got my log tables with me, but I happen to know that the logarithm of 1·8 is 0·255, so I guess the logarithm of 1·714 will be about 0·23 .
Therefore the pH of the solution will be 2 - 0·23 = 1·77
try this
Molecular weight of H2SO4 = 2+32+64 = 98
Concentration = [0.21]/98 in 025l = [0.21]*4/98 moles/l
=.00857
Assuming 100% ionisation and both H atoms lost - check on this
Conc H+ = .017 Moles/ Litre
Take Log = -1.77
Therefore pH = 1.77

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