Consider a 0.35 M solution of hyporbromous add (HOBr) in water. The equation for

Question

Consider a 0.35 M solution of hyporbromous add (HOBr) in water. The equation for the dissociation of this weak acid is HOBr (aq) H^+(aq) + OBr(aq) and K = 2.4 times 10^-9. What are the HOBr(aq), H^+(aq), and OBr^-(aq) concentrations at equilibrium? This problem is much easier to solve if you recognize that K is a very small number and so you can make the approximation that (HOBr(aq)] = 035-x ~ 0.35. Why is it okay to use this approximation? The pH of an acid can be calculated using pH = -log[H_3O^+] What is the pH of a 0.35 M solution of hypobromous acid? What Is the hydroinium ion concentration of a HOBr solution at pH 334? Now try a complete weak acid pH calculation Given that Ka for hypochlorous add is 2.9 times 10^-8 calculate the pH of 0.35M HOCl?

Explanation / Answer

Given the reaction is

HOBr <> H+ OBr-

Ka= [H+] [OBr-]/ [HOBr]

Initially : [HOBr] = 0.35M, [H+] = [OBr-] = 0

Let x= drop in concentration of [HoBr] to reach equilibrium

At Equilibrium [HoBr] =0.35-x   [H+] =[OBr-]=x

Ka= x2/ (0.35-x)= 2.4*10-9

Looking at the magnitude of Ka , it is reasonable to assume 0.35-x is approximately equal to 0.35

  Hence x2= 0.35*2.4*10-9

Hence x= 2.9*10-5, [H+] = [OBr-] = 2.9*10-5, [HOBr] =0.35-2.9*10-5 =0.3499

2. pH= -log (2.9*10-5)= 4.54

3. pH= 3.34, [H+] = 0.000457

4. As done in problem 1, Ka= x2/ (0.35-x)= 2.9*10-8

0.35-x can be approximated to 0.35

x2= 0.35*2.9*10-8, x= 0.0001007, pH= -log [0.0001007)=3.99

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