1A) A 37.6 mL sample of a 0.551 M aqueous hypochlorous acid solution is titrated
ID: 637178 • Letter: 1
Question
1A)
A 37.6 mL sample of a 0.551 M aqueous hypochlorous acid solution is titrated with a 0.242 M aqueous barium hydroxide solution. What is the pH after 18.6 mL of base have been added?
PH= ???
B)
What is the pH at the equivalence point in the titration of a 20.1 mL sample of a 0.441 M aqueous acetic acid solution with a 0.301 M aqueous barium hydroxide solution?
PH=???
c)
When a 21.1 mL sample of a 0.324 M aqueous hydrofluoric acid solution is titrated with a 0.338 M aqueous barium hydroxide solution, what is the pH after 15.2 mL of barium hydroxide have been added?
PH= ???
Explanation / Answer
1A) Ba(OH)2 + 2 HClO(aq) -----> Ba(ClO)2(aq) + 2 H2O(l)
1 mol Ba(OH)2 = 2 mol HClO
No of mol of HClO = 37.6*0.551 = 20.72 mmol
no of mol of Ba(OH)2 = 18.6*0.242 = 4.5 mmol
concentration of excess HClO = (20.72-4.5*2)/(37.6+18.6)
= 0.2085 M
pH = - log(H3O+)
= -log(0.2085)
= 0.68
B)
Ba(OH)2 + 2 ch3cooh(aq) -----> Ba(ch3coo)2(aq) + 2 H2O(l)
1 mol Ba(OH)2 = 2 mol ch3cooh
No of mol of ch3cooh = 20.1*0.441 = 8.864 mmol
no of mol of Ba(OH)2 = 8.864/2 = 4.432 mmol
volume of Ba(OH)2 consumed = 4.432/0.301 = 14.72 ml
Concentration of salt formed = 4.432/(201.+14.72) = 0.0205 M
pH = 7+1/2(pka+logC)
= 7+1/2(4.74+log0.0205)
= 8.526
c) Ba(OH)2 + 2 HF(aq) -----> Ba(F2)2(aq) + 2 H2O(l)
1 mol Ba(OH)2 = 2 mol HF
No of mol of HF = 21.1*0.324 = 6.84 mmol
no of mol of Ba(OH)2 = 0.338*15.2 = 5.14 mmol
no of mol of Ba(OH)2 reacted = 6.84/2 = 3.42 mmol
[OH-] = ((5.14-3.42)*2)/(21.1+15.2) = 0.095 M
pOH = -log(OH-)
= -log0.095
= 1.022
pH = 14-1.022 = 12.978
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