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Do you have the answer?Hey, can anyone help me with this question?
Calculate the pH of the resulting solution when 25.0mL of 0.750 mol/Lhydrochloric acid solution is added to 10.0 mL of 0.500 mol/L of barium hydroxide solution???
Thanks!!!!![]()
Do you still need help with it, or did you get it?yes it was a multiple choice question
options are:
a) 0.456
b) 0.602
c) 1.862
d) 2.058
answer was B in the end
ok, your moles for HCl is right but its (OH)2 in Ba(OH)2 so its 0.005 x 2 = 0.01.still need help. did you manage to work it out?
i got this far:
2HCl + Ba(OH)2 --> BaCl2 + 2H2O
I know no. of moles for HCl is 0.01875mol
no. of moles for Ba(OH)2 is 0.005mol
i know the HCl is in excess but where do i go from here?
yes it worked!! thankyou so much for your helpok, your moles for HCl is right but its (OH)2 in Ba(OH)2 so its 0.005 x 2 = 0.01.
Then work out how many moles of H+ are present after all the OH are neutralised (that's just the difference between them) which is 0.00875.
Then work out the concentration of that H+ by dividing 0.00875 by 0.025 + 0.01 (the total volume). Then do your logShould give you 0.602.
no worries, sorry for not getting back to you sooneryes it worked!! thankyou so much for your help![]()
Yeah, I wasn't being serious. Education should be free for all, not commercial like CSSA.I just think its stupid to try and stop people from sharing them.
+1Some of the multiple choice was a bit odd.
sameRelatively easy paper, how'd everyone else find it. I got stumped by the 3rd multiple choice lol.