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al Ar story Bookmarks People Window Help xWestern University, Ontario -x C Calculate The PH For Each of X php?id=278401 owta uivesty Gmail D uwo emaa Camp-Biology 11 Netflix Jump to Sapling Learning Calculate the phi for each of the following cases in the titration of 35 0 mL of 0.250 M KOHa. with 0.250 M HC(aq). with two decimal places (a) before addition of any HC (b) ater addision of 13.5 mL of HC (c) after addition of 245 mL of HC (d) after the adion of 35 0 mL of HCl (e) after the addition of 45.5 mL of HaO ) after the addition of 50.0 ml of HaD ToExplanation / Answer
a) concentration of KOH = 0.25 M
pOH = -log(OH-)
= -log0.25
= 0.6
pH = 14 - pOH
= 14-0.6
= 13.4
b) KOH(aq) + HCl(aq) ----> KCl(aq) + H2O(l)
No of mol of KOH taken = 35*0.25 = 8.75 mmol (higher)
No of mol of HCl taken = 13.5*0.25 = 3.375 mmol (lesser)
Molarity of mixer = M1V1-M2V2/V1+V2 (M1V1 > M2V2)
= (8.75-3.375)/(35+13.5)
= 0.111 M
pOH = -log(0.111) = 0.955
pH = 14-0.955 = 13.045
c)
KOH(aq) + HCl(aq) ----> KCl(aq) + H2O(l)
No of mol of KOH taken = 35*0.25 = 8.75 mmol (higher)
No of mol of HCl taken = 24.5*0.25 = 6.125 mmol (lesser)
Molarity of mixer = M1V1-M2V2/V1+V2 (M1V1 > M2V2)
= (8.75-6.125)/(35+24.5)
= 0.044 M
pOH = -log(0.044) = 1.36
pH = 14-1.36 = 12.64
d)
KOH(aq) + HCl(aq) ----> KCl(aq) + H2O(l)
No of mol of KOH taken = 35*0.25 = 8.75 mmol
No of mol of HCl taken = 35*0.25 = 8.75 mmol
both are equal,
so that, mixture is neutral, pH = 7
e) KOH(aq) + HCl(aq) ----> KCl(aq) + H2O(l)
No of mol of KOH taken = 35*0.25 = 8.75 mmol (lesser)
No of mol of HCl taken = 45.5*0.25 = 11.375 mmol (higher)
Molarity of mixer = M1V1-M2V2/V1+V2 (M1V1 > M2V2)
= (11.375-8.75)/(35+45.5)
= 0.0323
pH = -log(H3O+)
= -log(0.0323)
= 1.5
f) KOH(aq) + HCl(aq) ----> KCl(aq) + H2O(l)
No of mol of KOH taken = 35*0.25 = 8.75 mmol (lesser)
No of mol of HCl taken = 50*0.25 = 12.5 mmol (higher)
Molarity of mixer = M1V1-M2V2/V1+V2 (M1V1 > M2V2)
= (12.5-8.75)/(35+50)
= 0.0441
pH = -log(H3O+)
= -log(0.0441)
= 1.35
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