ITCREM 1010-Fall17-BRIDSON Activities and Due Dates 011/20/2017 11:55 PM 0349/10

Question

ITCREM 1010-Fall17-BRIDSON Activities and Due Dates 011/20/2017 11:55 PM 0349/10011/20/2017 12:27 PM Pret ] Calculator Periodic Table re Grade Question 14 of 25 Sapling Learning Map Forte titration of 55°nLof0.300 M NHs with 0500 M HCI at 25, of these points determine the relative pH at each O pH>7 (a) before the addition of any HCI O pH7 O PH 7 (b) after 33.0 mL of HCI has been addedO pH7 (c) after 43.0 ml of HCI has been added O pH-7 about us careers privacy policy serms of secomact us help

Explanation / Answer

Solution:- (a) NH3 is a weak base, so when no HCl is added to it then pH could be calculated making an ice table as..

NH3(aq) + H2O(l) <------> NH4+(aq) + OH-(aq)

I 0.300 0 0

C -X +X +X

E 0.300 - X X X

Kb = [NH4+][OH-]/[NH3]

1.8 x 10-5 = (X)2/(0.300 - X)

Since the Kb is very low so 0.300 - X could be taken as 0.300

1.8 x 10-5 = (X)2/0.300

X2 = 1.8 x 10-5 x 0.300

X2 = 5.4 x 10-6

X = 2.32 x 10-3

[OH-] = 2.32 x 10-3

pOH = - log 2.32 x 10-3

pOH = 2.63

pH = 14 - 2.63

pH = 11.37

So, first choice, pH > 7 is correct.

(b) Initial moles of NH3 = 0.0550 L x 0.300 mol/L = 0.0165 mol

moles of HCl added = 0.0330 L x 0.500 mol/L = 0.0165 mol

We have equal moles of weak base and strong acid so the solution would be acidic and the pH would be less than 7.

Third choice is correct, pH <7.

(c) When 43.0 mL of HCl are added to the NH3 then solution would have excess of acid and so the pH would be less than 7.

the right choice is the last one, pH<7.

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