Question

Ch. Ex. 84 - Calculate Concentrations from Hydrogen Sulfate Dissociation 2 attempts loft Check my work Be sure to answer all parts. Report problem What are the concentrations of HSO, SO2, and in a 0.11 M KHSO solution? (Hint: H2SO, is a strong acid; K, for HSO -1.3 x 10 ?) Hint eBook Print [HO] - [s0,?] - M M References

Answer 1

Solutions Given - concentration of KHSO4 = 0.11M Ka for 4500 = 1.3x10-2 The dissociation of KHSO4 is shown below- KHSO4 - Kt + HS05' Here 1 mal of KHS04 gives 1 mol of HSO4. Therefore, the concentration of KHSO will be same as the concentration of HS05. Therefore, the concentration of a sou will also be o.llM. Equation for the dissociation of itsoa is H505 = Sou + H Initial o.ll 0 0 change an th the a u Equilibrium 0.ll-n we know that - ka= [soa ?] [4] [Itsos

ka= () (2) (0.1l-) 1.3x102 = n² Ollon x² = 1.3410 2 (0.l1-2) 2 = 0.00143 – 0.0138 22 +0.0132 -0.00143 = 0 x= -0.013 .013) 2-4*1*(-0.00143) 2x 1 = -0.013✓ 5.88 9X10-3 a = -0.013+ 0.0767 2 malou -0.013 + 0.0767 -0.013-0.0767 2 a = (0.03185, or -0.04405 ) The concentration can not be regative,

Therefore only positive value is acceptable =+ 0.03185 M H+] = r = 0.03185 M (H+] = 0.03185 M ans (504°) = 0.03185 M ARS The concentration of H500 = 0.11-2 = 0.11-0.03185 = 0.07815 M TH000) = 0.07815M ADS Ans (HSOA] = 0.07 815 M. (5042] = 0.03185 m [H+] = 0.03185 M