1) What is the pH of 0.075 M CH3COOH? (Ka = 1.8 x 10-5) 2) What is the pH of 0.1

Question

1) What is the pH of 0.075 M CH3COOH? (Ka = 1.8 x 10-5)

2) What is the pH of 0.10 M NaCH3COO?

3) What is the pH of a solution containing 0.075 M CH3COOH and 0.10 M NaCH3COO?

4) What is the pH change that occurs when 0.010 mol HCl is added to 1.0 L of pure (i.e. unbuffered) water?  

pH of pure H2O =

pH of 0.010 M HCl =

    pH =

5) In contrast, what is the pH change that occurs when 0.010 mol HCl is added to 1.0 L of a buffer solution containing 0.10 M CH3COOH and 0.10 M NaCH3COO? (Assume no volume change. The Ka of CH3COOH = 1.8 x 10-5.)

pH before adding the acid:

pH after adding the acid:

Explanation / Answer

Acetic acid is weak acid and ionizes as CH3COOH-----> CH3COO- + H+

Ka= [CH3COO-] [H+]/[CH3COOH]

              CH3COOH                           CH3COO-                  H+

Initial      0.075                               0                               0

Drop         -x                                   x                                x

Eq          0.075-x                          x                                    x

Ka= x2/(0.075-x)= 1.8*10-5, when solved using spread sheet of excel,

The value of [H+] =0.00115, pH= 2.94

2.

2. CH3COONa is salt formed from weak acid (CH3COOH) and strong base (NaOH)

CH3COO-+ H2O--------->CH3COOH + OH-

Kb= [CH3COOH] [OH-]/[CH3COO-]= Kw/Ka= 10-14/1.8*10-5=5.6*10-10

                 CH3COO-                    CH3COOH                  OH

Initial      0.1                                   0                               0

Drop         -x                                   x                                x

Eq          0.1-x                          x                                    x

Hence x2/(0.1-x)= 5.6*10-10, when solved using excel,

[OH-] =7.5*10-6, POH= 5.12, pH= 14-5.12=8.88

3.

pH= pKa + log [conjugate base/acid]

PKa of acetic acid =-log (1.8*10-5)=4.74

pH =4.74+ log (0.1/0.075) =4.86

4. Pure water pH= 7

when 0.010mole of HCl is added to 1 L of water, concentration of HCl =0.010/1=0.010M

HCl ionizes completely in water as HCl -------> H+ +Cl-

pH= -log (0.01)=2, change in pH= 7-2= 5

5. before additino of HCl, pH= pKa + log [conjugate base / acid] =4.74

b) Moles of CH3COONa in 1L buffer = 0.1*1=0.1 moles, HCl added =0.010

CH3COONa+ HCl---------->CH3COOH+NaCl , 1 mole of CH3COOH requies 1 mole ofHCl. Since HCl is 0.010 mole scompared to 0.1M CH3COONa, since HCL is the limiting reactants, all the HCl is consumed. moles of CH3COOh formed = 0.1+0.01= 0.11 , moles of CH3COONa= 0.1-0.01=0.09

pH= 4.74 + (0.09/0.11) =4.65

drop in pH= 4.74-4.65=0.09

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