Question

Calculate the pH for each case in the titration of 50.0 mL of 0.240 M HClO(aq) with 0.240 M KOH(aq). Use the ionization constant for HCIO. What is the pH before addition of any KOH? pH = 4 What is the pH after addition of 25.0 mL KOH? pH = 7.4 What is the pH after addition of 40.0 mL KOH? Question Source: MRG-General Chemistry | Publisher University Science help

I just need help with the 40.0 mL pH

Answer 1

50 ml of 0.24 Ka of telo 23.0xco m Huo (aq) Helo is a weah acid. Helo Ht + con I 0.24 m D 0 с ta tn E N N ka 3x10 ny 6.24 ) Now x is x is very small 30 (0.24-2) = 0.24 IH) [co] THRO) nxa 0.24 0.72x10 - 8.485x105 1447 8.485x105 m pH = -log[**] - log [8-485x16) pH = 4.07 Ansy of kot pH of the solution - 4.07 before addition

50 ml of 0.24 m Ho 50 ml 8.485810 m. at ion -5 moles of Ht on a 5008 1485X10 1000 -5 0424 x 10 ondle 40 ml 0.24 m Kolt is added Kot is a trong base so complete dissociation is There KOH - kt+ oH of oH sons 40 x 0.26 1000 0.0006 mole moles 18 moles of excess or ions - (090096 - 04424X16 20100959 ondt Total solution (50ml + 40 m) 90 ml so [OH) = 0.00959 90 0.106 m log [OH-] 0.97 pH = 14-polt Boy 14-097) - 13.03 Anse after addition of of 40 ml KOH PH2 13.63 X(ooo 2 Potta - log [o. 106) 2