Question

2.5 g of sulphuric acid is present in 250 cc of solution, what is its ph

Answer 1

molar mass of sulphuric acid is 1*2 + 32+ 4*16 => 98 grams per mole.

So, 2.5 grams of H2SO4 is 2.5/98 => 0.0255 moles

1 mole of H2SO4 furnishes 2 moles of of H+ in aqueous state.

So, 0.0255 moles of H2SO4 will furnish 2*0.0255 => 0.051 moles of H+ .

Volume of solution is 250 cc or 0.25 litres.

Molarity of H+ ion in the solution= (moles of H+)/ litres of solution

=> 0.051/0.25

=> 0.051*100/25

=>0.051 * 4 => Molarity of H+ ions= 0.204 M

So, PH of solution = -log (base 10) H+ concentration.

=> - log (base 10) 0.204   (from calculator log (base 10) 0.204 = -0.69)

=> -1 * -0.69

=> 0.69

Hence the PH of the solution will be 0.69 (approximately)