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CHEMISTRY HELP - acids and bases

can someone please help me with this q,
calculate the ph of the solution formed when 15cm^3 of 0.12moldm^-3 naoh has been added to 25cm^3 of 0.0566moldm^-3 of a weak monoprotic acid hx

in part (a) it did mention that hx has a pka of 4.73 at 25 degrees, but idk if thats relevant here
Reply 1
you might want to check the answer key cus this could be completely wrong:
NaOH + HX = NaX + H2O
Na+ is spectator ion so:
X- + H20 = HX+ OH-
mols of NaOH= (15x10^-3)(0.12)= 1.8x10^-3 mols
molsHX= C x V= 1.415x10^-3 mols (so limited reagent meaning mols of NaX= mols X-= 1.415x10^-3 mols

pKw=14= pKa+ pKb so pKb = 14- 4.73= 9.27 so Kb= 10^-9.27
and then you j do an ice table for the second reaction to fill in the formula for Kb and then solve for conc. of OH- (which is the same as conc. of HX). The conc of X- is mols/ (25+15)= 0.035375 mols/dm^3
Kb= ([OH-]x[HX])/[X-] = x^2/ 0.035375 = 10^-9.27, solve for x to find that [OH-]= 8.822x 10^-7 mol/dm3 so pOH= 6.05 and then pH= 14-6.05= 7.95
i hope this helps, i did this topic a few months ago so there might be some mistakes but i think this is the general way to do it? sorry if its wrong
Reply 2
Original post by ABCDE200
you might want to check the answer key cus this could be completely wrong:
NaOH + HX = NaX + H2O
Na+ is spectator ion so:
X- + H20 = HX+ OH-
mols of NaOH= (15x10^-3)(0.12)= 1.8x10^-3 mols
molsHX= C x V= 1.415x10^-3 mols (so limited reagent meaning mols of NaX= mols X-= 1.415x10^-3 mols

pKw=14= pKa+ pKb so pKb = 14- 4.73= 9.27 so Kb= 10^-9.27
and then you j do an ice table for the second reaction to fill in the formula for Kb and then solve for conc. of OH- (which is the same as conc. of HX). The conc of X- is mols/ (25+15)= 0.035375 mols/dm^3
Kb= ([OH-]x[HX])/[X-] = x^2/ 0.035375 = 10^-9.27, solve for x to find that [OH-]= 8.822x 10^-7 mol/dm3 so pOH= 6.05 and then pH= 14-6.05= 7.95
i hope this helps, i did this topic a few months ago so there might be some mistakes but i think this is the general way to do it? sorry if its wrong

thank you so much can i just ask u, yk in the second part, where it says pKw i dont get what exactly you've done there
Reply 3
Original post by lol_124
thank you so much can i just ask u, yk in the second part, where it says pKw i dont get what exactly you've done there

its a rule we learned that the water constant (Kw) is the Ka of an acid multiplied by the Kb of its conjugate base, so w log laws you get the formula pKw= pKaxpKb and since the temperature is room temp the pKw is 14 so u can j find the pKb using that
Reply 4
oops meant + not x w the pKa n pKb

Original post by ABCDE200
its a rule we learned that the water constant (Kw) is the Ka of an acid multiplied by the Kb of its conjugate base, so w log laws you get the formula pKw= pKaxpKb and since the temperature is room temp the pKw is 14 so u can j find the pKb using that
Reply 5
Original post by ABCDE200
its a rule we learned that the water constant (Kw) is the Ka of an acid multiplied by the Kb of its conjugate base, so w log laws you get the formula pKw= pKaxpKb and since the temperature is room temp the pKw is 14 so u can j find the pKb using that

ah ok that makes sense,
thank u so much x

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