Sulfuric acid (H2SO4) is a strong acid, which dissociates when dissolved in water according to the following equation:

H2SO4 (aq) → 2H+ (aq) + SO42-(aq)

 

A 0.21 g sample of sulfuric acid is dissolved completely in sufficient water to make 0.25 litre of final solution. Calculate the hydrogen ion concentration (in mol l−1) in this solution. Give your answer in scientific notation to an appropriate number of significant figures. Remember to show the successive steps in the calculation, and to explain your reasoning.

What is the pH of the sulfuric acid solution (to the nearest whole number)?

This forum does not do your homework for you. This forum will help you along the way, and show you what mistakes you may have made and/or confirm the correct answer.

 

With that, please show what work you have done thus far, and where in that work you have done you are having problems.

i agree with bignose

first you must try yourself

helping doesnot mean to make you idle

dont make yourself sluggish by jamming your brain try it on your own and

show that you have done something

dont worry if u cant we will hellp you solve it out surely dont worry:-)

ok, can we wait for his reply next, before we read the riot act to the guy yet again?

This is what I have done so far but Im not sure if Im on the right track

 

The molarity of the H2SO4 initially is 0.21g/98mols

 

= 0.00214 and that in 0.25 L = 0.00285714285 = 0.0086 mols/L.

 

 

The first ionization will result in H^+ of 0.0086 mols/L.

Consider a 0.1 mol dm-3 solution of sulphuric acid. The equilibria in this solution are:

 

H2SO4(aq) + H2O(l) H3O+(aq) + HSO4- (aq) K1 = very large

 

H2O(l) + HSO4- (aq)  H3O+(aq) + SO42-(aq) K2 = 0.01 mol dm-3

 

A 0.1 mol dm-3. solution of sodium hydrogen sulphate NaHSO4 which contains only the second ionisation shown above has a pH of 1.57. This corresponds to a hydrogen ion concentration of 0.027 mol dm-3. Thus it seems that 0.1 mol dm-3 sulphuric acid should have a hydrogen ion concentration of (0.1 + 0.027) = 0.127 mol dm-3

 

Consider a 0.1 mol dm-3 solution of sulphuric acid. The equilibria in this solution are:

 

H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4- (aq) K1 = very large

 

H2O(aq) + HSO4- (aq)  H3O+(aq) + SO42-(aq) K2 = 0.01 mol dm-3

 

If the concentration of the hydrogen ions from the second ionisation only is x, then the concentration of HSO4- ions will be 0.1 - x and the total concentration of hydrogen ions from both ionisations will be

0.1 + x. Since it is the total hydrogen ion concentration that appears in the evaluation of K2, we have:

K2 = [H3O+] [sO42-] = (0.1 + x) x = 0.01 mol dm-3

[HSO4-] (0.1 – x)

Since a concentration must be a positive quantity the negative root makes no physical sense; thus

x = 8.45 x 10-3 mol dm-3.

The total hydrogen ion concentration is therefore 0.10845 mol dm-3 = (0.109 mol dm-3)

PH of 0.96