Question

TUTOR STEP The equilibrium concentrations can then be calculated from the value of x. Enter them here. H2O(l) HB+ (aq) + 0.15 0 B(aq) + Initial (mol/L) Change (mol/L) -8.7 x 10-5 Equilibrium (mol/L) 0.15 – 8.7 x 10-5 OH(aq) 0 +8.7 x 10-5 8.7 x 10-5 +8.7 x 10-5 8.7 x 10-5 pOH = -log[OH ]. Use this to calculate pH using pH = 14.00 - POH. [B] = 0.149 mol/L [HB]= mol/LX [OH'] = mol/L X pH = 9.94 Recheck Next (9 of 9) 3rd attempt

Answer 1

The reaction ý given B(aq) + H2O (L) = initial → 0.15 (M) here - 49 He® (as) + oH® (29) OM OM) ober 78.7x10 m) +8.7x10 ) Change → -8.7x10 cm Concentration at equlibrium - (0:15-8:7x105) (M) 10t8.7x105) (M) (ot87 x605) (M) B = : equilibrium concentration of (0.15-8.7x10-5 ) M = 0.1499 (M). = 0.1499 modth The concentration of Oho at equilibrium = 8.7x10-5 M) = 8.7x10-5 mol/L As we know, poh = -logio [OH-] [OH-] = concentration of oHo. pot _ - log (8.7x 105) = 4.0604 we know pH + polt = 14 "

pH = 145 POH = 14 - 4. 0604 PHOTO Gismon dot + cope) The concentration of [BH] at equlibrium is also 8.7x10 (M) [BH = 8.7*105 mol/L at equlibrium.