# CHEMISTRY HELP - acids and bases

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
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
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
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
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
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