find the pH of 300 mL of an aqueous of HI 0.38 M with 270 mL of another solution of potassium hydroxide 0.646 M.
answer key: 13.025
but i keep finding 13.704 for a basic solution
can someone check it thanks
hi
the reaction is SA + SB = WCB + WCA
with HI + KOH = IOH + HK
n HI = 0,38 * 0,3 = 0,114
n KOH = 0,27 * 0,646 = 0,174
after reaction we are left with 0,174-0,114=0,06 mol of KOH in excess
so we also have 0,06 mols of HO⁻ ions in excess in the solution
C = 0,06/0,57 = about 10⁻¹ mol/L
pOH = -log10[10⁻¹] = 1
pH = 14-1 = 13