HSC Chemistry Question Thread

[jakesilove]

Hey,
would someone please be able to help me solve this question:

what is the pH of a mixture of 20.0 mL of 0.102 mol L^-1 barium hydroxide solution and 40.0 mL of 0.150 mol L^-1 hydrochloric acid diluted to a final volume of 100 mL

Check out my answer here, which is the same question but with different numbers. Read the next few posts as well, because I forgot to include the 1:2 ratio initially

[ProfLayton2000]

Hey,
would someone please be able to help me solve this question:

what is the pH of a mixture of 20.0 mL of 0.102 mol L^-1 barium hydroxide solution and 40.0 mL of 0.150 mol L^-1 hydrochloric acid diluted to a final volume of 100 mL

The reaction for this reaction is Ba(OH)2(aq) + 2HCl(aq) --> BaCl2(aq) + H2O(l)
n(Ba(OH)2) = c*v = 0.102 * 0.02 = 0.00204mol
n(HCl) = c*v = 0.150*0.4 = 0.006mol
Because Ba(OH)2 and HCl are supposed to react in a 1:2 ratio, we can see that Ba(OH)2 is the limiting reagent.
Moreover, we can deduce that because all of the Ba(OH)2 will be used (0.00204mol), it can calculated that n(HCl used) = n(Ba(OH)2) * 2 = 0.00408mol.
Next, we show that: n(HCl) = 0.006 -  0.00408 = 0.00192mol, hence there are 0.00192 moles of HCl in excess.
Next, we find the concentration of HCl in the final mixture: n/v = 0.00192/(0.1) = 0.0192 molL-1
HCl is monoprotic so we can say that therefore [H+] = 0.0192molL-1
And pH = -log[H+] = log(0.032) = 1.72 (3sf)

Can someone check answer including sig figs are right

[jakesilove]

The reaction for this reaction is Ba(OH)2(aq) + 2HCl(aq) --> BaCl2(aq) + H2O(l)
n(Ba(OH)2) = c*v = 0.102 * 0.02 = 0.00204mol
n(HCl) = c*v = 0.150*0.4 = 0.006mol
Because Ba(OH)2 and HCl are supposed to react in a 1:2 ratio, we can see that Ba(OH)2 is the limiting reagent.
Moreover, we can deduce that because all of the Ba(OH)2 will be used (0.00204mol), it can calculated that n(HCl used) = n(Ba(OH)2) * 2 = 0.00408mol.
Next, we show that: n(HCl) = 0.006 -  0.00408 = 0.00192mol, hence there are 0.00192 moles of HCl in excess.
Next, we find the concentration of HCl in the final mixture: n/v = 0.00192/(0.1) = 0.0192 molL-1
HCl is monoprotic so we can say that therefore [H+] = 0.0192molL-1
And pH = -log[H+] = log(0.032) = 1.72 (3sf)

Can someone check answer including sig figs are right

Looks perfect!

[ProfLayton2000]

Does it matter whether we take bromine water to be Br2 or HOBr?

[wesadora]

Does it matter whether we take bromine water to be Br2 or HOBr?

Nah, both are fine (HOBr is 'more right' but they don't care cuz we're "just" HSC chem bahahaha). I just do Br2 for simplicity (but aqueous ofc)

[Cindy2k16]

Could they ask us to write the chemical formula/ draw the structural formula for the polymer and/or monomer of our chosen biopolymer?
TIA

[jakesilove]

Could they ask us to write the chemical formula/ draw the structural formula for the polymer and/or monomer of our chosen biopolymer?
TIA

I'm fairly sure you don't need to draw it, or know the empirical formulas, but you need to be able to state the name of the polymer and associated monomer

[Cindy2k16]

I'm fairly sure you don't need to draw it, or know the empirical formulas, but you need to be able to state the name of the polymer and associated monomer

oh thank goodness. Thanks!

[noonedoesnt]

I got 4.65g, which is closest to B, which is the correct answer. Anyone have an idea why its 5.30g?

[noonedoesnt]

oh thank goodness. Thanks!

I can't recall the question, maybe from a trial paper, but it asked to draw our biopolymer. Mine was PHBV, which is difficult to draw and I didn't know what to do at the time lol

[jakesilove]

I can't recall the question, maybe from a trial paper, but it asked to draw our biopolymer. Mine was PHBV, which is difficult to draw and I didn't know what to do at the time lol

I seriously doubt that it is even assessable in an HSC exam

[noonedoesnt]

I seriously doubt that it is even assessable in an HSC exam

I would really hope not!

[jakesilove]

I got 4.65g, which is closest to B, which is the correct answer. Anyone have an idea why its 5.30g?

If you take the equation to be 1:1, then you get 0.05 moles of each substance, right? The molar mass of sodium carbonate is 106, so 0.05*106=5.3g

[noonedoesnt]

If you take the equation to be 1:1, then you get 0.05 moles of each substance, right? The molar mass of sodium carbonate is 106, so 0.05*106=5.3g

Oh thanks, stupid error. Calculated wrong molar mass, used molar mass of NaHCO3... thanks

[jakesilove]

Oh thanks, stupid error. Calculated wrong molar mass, used molar mass of NaHCO3... thanks

Fair enough! Shit happens, and you got the right answer anyway