Chemistry Question (1 Viewer)

[Reikira]

Calculate the pH of the resulting solution when 25.0mL of 0.750mol L HCl solution is added to 10ml of 0.500mol L barium hydroxide solution.

Cant figure this one out :<

 

[BadMeetsEvil]

Calculate the pH of the resulting solution when 25.0mL of 0.750mol L HCl solution is added to 10ml of 0.500mol L barium hydroxide solution.

Cant figure this one out :<

is the answer 2.06?

Edit: the answer is 0.60

 

[Dylanamali]

The reaction that takes place: 2HCl (aq) + Ba(OH)2 (aq) -> Ba(Cl)2 (aq) + H2O (l)

First Step, they give you the concentration and volume of both solutions, so you can work out the number of mols of each reactant, using n = C x V

therefore, n. of HCl = 0.0250 x 0.750 = 0.01875 mols
n. of Ba(OH)2 = 0.01 x 0.500 = 0.005 mols

applying molar ratios, HCl : Ba(OH)2 = 2:1
therefore n HCl reacted = 2 x 0.005 = 0.01 mols
therefore HCl in excess is 0.01875 - 0.01 = 0.00875 mols

C HCl = n/V = 0.00875/0.025 + 0.01 (addition of volume of Barium hydroxide to HCl)
= 0.25 M

ph = -log [H+] = -log(0.25) = 0.602 (3 sig figs)

 

[BadMeetsEvil]

The reaction that takes place: 2HCl (aq) + Ba(OH)2 (aq) -> Ba(Cl)2 (aq) + H2O (l)

First Step, they give you the concentration and volume of both solutions, so you can work out the number of mols of each reactant, using n = C x V

therefore, n. of HCl = 0.0250 x 0.750 = 0.01875 mols
n. of Ba(OH)2 = 0.01 x 0.500 = 0.005 mols

applying molar ratios, HCl : Ba(OH)2 = 2:1
therefore n HCl reacted = 2 x 0.005 = 0.01 mols
therefore HCl in excess is 0.01875 - 0.01 = 0.00875 mols

C HCl = n/V = 0.00875/0.025 + 0.01 (addition of volume of Barium hydroxide to HCl)
= 0.25 M

ph = -log [H+] = -log(0.25) = 0.602

yeah correct. Oops forgot to find the concentration of H+ in my answer.. Just went straight to -log(0.00875). My bad

 

[Bobbo1]

The reaction that takes place: 2HCl (aq) + Ba(OH)2 (aq) -> Ba(Cl)2 (aq) + H2O (l)

First Step, they give you the concentration and volume of both solutions, so you can work out the number of mols of each reactant, using n = C x V

therefore, n. of HCl = 0.0250 x 0.750 = 0.01875 mols
n. of Ba(OH)2 = 0.01 x 0.500 = 0.005 mols

applying molar ratios, HCl : Ba(OH)2 = 2:1
therefore n HCl reacted = 2 x 0.005 = 0.01 mols
therefore HCl in excess is 0.01875 - 0.01 = 0.00875 mols

C HCl = n/V = 0.00875/0.025 + 0.01 (addition of volume of Barium hydroxide to HCl)
= 0.25 M

ph = -log [H+] = -log(0.25) = 0.60 (3 sig figs)

if it was 3 significant figures it would be 0.600 or 6.00 x 10^-2 right?

 

[b3kh1t]

if it was 3 significant figures it would be 0.600 or 6.00 x 10^-2 right?

yep well done

 

[wb47]

I am kinda confused with this as well, why can't you divide n(HCl) by 2 instead of 2n[Ba(OH)2]?
Thanks

 

[b3kh1t]

I am kinda confused with this as well, why can't you divide n(HCl) by 2 instead of 2n[Ba(OH)2]?
Thanks

because when HCl ionises only one hydrogen ion is produced, but when Ba(OH)2 ionises 2 hydroxide ions are produced. Not sure if that will answer your question, let me know if it doesn't

 

[BadMeetsEvil]

I am kinda confused with this as well, why can't you divide n(HCl) by 2 instead of 2n[Ba(OH)2]?
Thanks

you multiply the n of Ba(OH)2 by 2 because for one mole of barium hydroxide, you release 2 moles of hydroxide ion. and you're trying to find the amount of moles of hydroxide ions because that's what reacts with H+. and you can see that theres more h+ than oh- therefore oh- is the limiting reagent (gets all used up) and you base your stoichiometry on the amount of moles of oh-

 

[wb47]

oh I see if it was THEORETICALLY 2HCl + BaOH the acidic mol in excess would just be 0.01875 - 0.005?

Thanks

 

[b3kh1t]

oh I see if it was THEORETICALLY 2HCl + BaOH the acidic mol in excess would just be 0.01875 - 0.005?

Thanks

yep basically

 

[wb47]

but then I don't see why Dylanamali used the HCl: Ba(OH)2 molar ratio in his calculations, isn't it redundant as the 2:1 is the ratio required fort he neutralisation reaction?

Thanks +rep.

EDIT: oooooh now i see it the 0.005 is the limiting reagent and he got the molar used up for the HCl with the ratios.
EDIT2: Wait i might be wrong...

 

[b3kh1t]

but then I don't see why Dylanamali used the HCl: Ba(OH)2 molar ratio in his calculations, isn't it redundant as the 2:1 is the ratio required fort he neutralisation reaction?

Thanks +rep.

EDIT: oooooh now i see it the 0.005 is the limiting reagent and he got the molar used up for the HCl with the ratios.
EDIT2: Wait i might be wrong...

Yes but since 2 moles of HCl are used for 1 mole of Ba(OH)2, then the limiting agent will be not 0.005 moles of HCl but 2x0.005=0.01 moles of HCl

 

[BadMeetsEvil]

Yes but since 2 moles of HCl are used for 1 mole of Ba(OH)2, then the limiting agent will be not 0.005 moles of HCl but 2x0.005=0.01 moles of HCl

the limiting agent is Ba(OH)2 so it is 0.01 moles of hydroxide ions

 

[b3kh1t]

the limiting agent is Ba(OH)2 so it is 0.01 moles of hydroxide ions

Oops sorry wrong way round but thank you

 

[wb47]

i am soo sorry for dragging on but if 0.005 is the limiting agent, and so the mol of HCl is 0.01, how does that relate to the ionisation of Ba(OH)2 having 2 OH-s is that already accounted for in the equation?

Thanks again.

 

[BadMeetsEvil]

i am soo sorry for dragging on but if 0.005 is the limiting agent, and so the mol of HCl is 0.01, how does that relate to the ionisation of Ba(OH)2 having 2 OH-s is that already accounted for in the equation?

Thanks again.

0.005 is the moles of Ba(OH)2 and when that dissociates.. It breaks down to 2 OH so that means that 0.01 OH will be released. And from this, you compare the moles of H+ and OH. And we can see that OH has less moles than H which makes OH the limiting reagent.

 

[b3kh1t]

i am soo sorry for dragging on but if 0.005 is the limiting agent, and so the mol of HCl is 0.01, how does that relate to the ionisation of Ba(OH)2 having 2 OH-s is that already accounted for in the equation?

Thanks again.

Sorry but I will start from the beginning and go through, then if there is still anything you need don't bee afraid to ask



now number of moles of HCl = 0.025x0.75= 0.01875 moles
number of moles of Ba(OH)2 = 0.01x0.5= 0.005 moles

For every mole of Ba(OH)2, 2 moles of HCl is consumed
so 0.005 mol of Ba(OH)2 consumed 2x0.005 mol of HCl.
therefore 0.01 mole of HCl is used and the remaining HCl is 0.01875-0.01 = 0.00875 moles of H+ remain
therefore concentration of H+ ions is C= n/V = 0.00875/(0.025+0.010) = 0.25
then find the ph by -log(0.25)

another method is:
number of moles of HCl = 0.025x0.75= 0.01875 moles
therefore number of moles of H+ produced is 0.01875 moles

number of moles of Ba(OH)2 = 0.01x0.5= 0.005 moles
therefore number of OH- produced is 2x0.005 = 0.01 moles (this is because each molecule of Ba(OH)2 has 2 OH-)

thus the 0.01 moles of OH- will neutralise 0.01 moles of the H+, therefore 0.01875-0.01=0.00875 moles of H+ remaining
therefore concentration of H+ ions is C= n/V = 0.00875/(0.025+0.010) = 0.25
then find the ph by -log(0.25)

 

[BadMeetsEvil]

Sorry but I will start from the beginning and go through, then if there is still anything you need don't bee afraid to ask



now number of moles of HCl = 0.025x0.75= 0.01875 moles
number of moles of Ba(OH)2 = 0.01x0.5= 0.005 moles

For every mole of Ba(OH)2, 2 moles of HCl is consumed
so 0.005 mol of Ba(OH)2 consumed 2x0.005 mol of HCl.
therefore 0.01 mole of HCl is used and the remaining HCl is 0.01875-0.01 = 0.00875 moles of H+ remain
then find the ph by -log(0.00875)

another method is:
number of moles of HCl = 0.025x0.75= 0.01875 moles
therefore number of moles of H+ produced is 0.01875 moles

number of moles of Ba(OH)2 = 0.01x0.5= 0.005 moles
therefore number of OH- produced is 2x0.005 = 0.01 moles (this is because each molecule of Ba(OH)2 has 2 OH-)

thus the 0.01 moles of OH- will neutralise 0.01 moles of the H+, therefore 0.01875-0.01=0.00875 moles of H+ remaining
then finf the pH by -log(0.00875)

0.00875/volume. You want the concentration of H+. That's what you put in -log(H+)

 

[b3kh1t]

0.00875/volume. You want the concentration of H+. That's what you put in -log(H+)

Yep sorry was doing another question at the same time and did not realise, but as long as the molar ratio and such is understandable. Thanks