Sunday
May 19, 2013

# Homework Help: Chemistry

Posted by SOS mature student on Sunday, February 18, 2007 at 4:53pm.

Read earlier posts on this question and still find my self confused.
The question being H2SO4(aq) -> 2H^+ (aq) + SO4^-2 (aq). A 21g sample of sulfric acid dissolved completely in sufficient water to make a 0.25 l of final solution. Calculate hydrogen ion concentration in mols per litre. I have to show successive steps. And then determine the pH of the sulfric acid.

Going by the learning objectives laid out for the question I have found the concentration for one litre. So, i have divided grams by molar mass which is 0.84/98= 0.0086 to 2 sig figs.
Looking at a similar question in my text book they seem to be suggesting that because the formula when shows that for every molecule of sulfric acid dissolved, 2 hydrogen ions are formed. That I should, 2 X 0.0086 to get 0.017 mols per litre, 1.7 x 10^-2 2 sig figs, don't know if that is the right way to go about it. I am not sure how to determine the compound's pH.We are not expected to calculate ionizations as laid out in earlier post K2 etc not tounched on anything like that, it has to be kept basic. I don't understand from earlier post what pH = log(H^+) means? Haven't studied for a very long time, too long to mention so please be gentle with me. xxx

You question is a little confusing, too. You have 21 g H2SO4 in the first sentence; therefore, 21g/molar mass of 98 = ?? I don't know where you obtained the 0.84 g.

Concerning 2H^+ per mol H2SO4, that is ALMOST but not quite true. If you are just starting the course and haven't covered anything about k2 for H2SO4, then I would simply multiply by 2 as you suggested in your post.

As for pH, that is a scale for (H^+) and pH = - log(H^+) by definition.
Calculate (H^+), take the log (base 10) and place a negative sign in front of it.

I will try to answer more after you clarify the question.

I may of got confused, so I will start from begining.The whole question is as follows
H2SO4(aq)-> 2H^+(aq)+SO4^-2
A sample of sulfuric acid is dissolved completely in sufficient water to make 0.25 litre of final solution.Calculate the hyrogen ion concentration in mol litre^1 Give your answer in scientific notation to appropriate number of sig figs. Show successive steps in the calculation and explain reasoning.

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

I think that the way the text is teaching we must assume that the two hydrogen ionizations contribute the same.

Molar mass = (2 x 1.01)+ 32.1 + (4 x 16.0) = 98.12 = 98g to 2 sig figs. Therefore 0.21/98 = 0.0021mols per litre to 2 sig figs. I was following your previous post which gave 0.00857
mols per litre. The only way I could match this figure was to convert concentration to a litre 0.21g/250 cubic cms which gave me 0.00857 mols/l or 0.0086 to 2 sig figs. Because there are two H^+ in formula I, 2 x 0.0086 to get 0.017 mols/l 1.7 x 10^-2 2 sig figs, perhaps I have been doing it all wrong.

So, perhaps you could just talk me through this as simply as you can from the begining so i can fully understand, the more I try to understand the more lost I become. Please bear in mind I don't know what you mean by -log (base 10)to calculate H^+, I need it explained in simpler terms,so I can understand where I am going, haven't done maths for years, the text doesnt use the term logarithm. Been coping with a lot of the maths in the course but have hit problems since I started the chemistry section. I would be so grateful, feeling so frustrustedand depressed at my own inability to grasp this.

OK. We'll just forget about the k2 for H2SO4 and concentrate on there being two equal H^+ in H2SO4.
H2SO4 ==> 2H^+ + SO4^=

Is that 0.21 g H2SO4 in enough water to make 0.25 L? If that is right, then we have 0.21/98 mols H2SO4 and that in 0.25 L is 0.00214 mols/0.25L =0.00857 mols/L or 0.00857 M H2SO4.

(H^+) = 2 x 0.00857 M = 0.01714 M or 0.01714 mols/L. In scientific notation that would be 1.714 x 10^-2 mols/L and to two s.f. it would be 1.7 x 10^-2 mols/L. Note that I carry at least one extra s.f.(actually I carried more than that in the 0.01714 number) during all my calculations and round at the end. Some purists don't agree with that but as a practicing analytical chemist I found that rounding at each step often (usually) leads to rounding errors and rounding only once, instead of several steps, reduces that risk. (In real practice, what I do is as follows: I punched in 0.21 in the calculator, divided by 98 to obtain 0.00214----actually the number is 0.002142857 etc etc---but I DON'T erase that value. I just leave it in the calculator with all of those extra numbers, then divide by 0.25L to obtain 0.008581428 etc etc---again, I leave that number in the calculator. I multiply by 2 to obtain 0.0171428 etc etc--I don't remove that number either. Leave it in the calculator.

Now to convert to pH, there is no other way to do it than to use the conversion of pH = -log(H^+). If you are unfamiliar with that now is a good time to become familiar with it. With the 0.01714 number still in the calculator, look for the "log" button. There are two logarithm scales on most calculators, the natural logarithm (base e), abbreviated ln, and the logarithm base 10, abbreviated log. Punch the log button and up will pop -1.7659. I copy the -1.7659 number down. In the formula it looks this way.
pH = -log(H^+) = -log(0.01714) = -(-1.7659) = +1.7659 which I would round to 1.77. I hope this clears things up for you.

If you reply to this response, it may be best to start as a new question for this "old" question is on page 2 of the board today and it will continue to move to page 3, 5, etc as time goes on. It's hard to follow up on them when they get that far away from the front of the list.

Hi SOS mature student.

You are obviously suffering with s103 as I am. Having read page 79 of block 8, I think the answer is to use the power of ten as the PH which gives an answer of 2.

No sign of log etc and don't want anyone to see where answer came from.

No one has answered this question yet.

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