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You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will the ph of this solution be after the addition of 20.0 ml of 1.00 m naoh solution? [ka = 1.8 x 10-5]

Respuesta :

superman1987
First, we should get moles acetic acid = molarity * volume

                                                                =0.3 M * 0.5 L

                                                                =  0.15 mol

then, we should get moles acetate = molarity * volume

                                                           = 0.2 M * 0.5L

                                                           = 0.1 mol

then, we have to get moles of OH- which added:

moles OH- = molarity  * volume

                   = 1 M    * 0.02L

                  = 0.02 mol

when the reaction equation is:


                 CH3COOH  +  OH-  → CH3COO-   +  H2O


moles acetic acid after adding OH- = (0.15-0.02) 
                                             
                                                            =  0.13M                                       

moles acetate after adding OH- =  (0.1 + 0.02)

                                                      =   0.12 M

Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

                        = 0.13mol / 0.52L

                       = 0.25 M

and [acetate ) = 0.12 mol / 0.52L
 
                        = 0.23 M

by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

       = 4.7

So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

         = 4.7 + ㏒(0.23/0.25)

        = 4.66
abiolataiwo2015

The ph of this solution after the addition of 20.0 ml of 1.00 m naoh solution is 4.71.

First step is to calculate the dissociation equilibrium at 25 degrees Celsius  

CH3COOH⇆ H+  + CH3COO-  

Ka=1.8×10^-5= [H+]  × [CH3COO-] /[CH3COOH]

Second step is to simply the expression since both  buffer components are in the same final volume

Ka= [H+] × nCH3COO-/nCH3COOH

Third step is for the acetic acid to neutralized the strong base (NaOH) so as to create more acetate ion:

NaOH+CH3COOH→H2O+Na+ +CH3C00-

Fourth step is to calculate the weak species molar quantities

nCH3COOH=(0.5000L×0.30M)-(0.0200L×1.00M)

nCH3COOH=0.15-0.02

nCH3COOH=0.13mol

nCH3COO=(0.5000L×0.20M)+(0.0200L×1.00M)

nCH3COO=0.1+0.02

nCH3COO=0.12mol

Fifth step is to determine the new hydrogen ion molarity

1.8×10^-5= [H+] × 0.13mol/0.12mol

[H+] =2.0×10^-5M

Sixth step is to find the pH

pH=-log[H+]

pH=-log(2.0×10^-5)

pH=4.71

Inconclusion the ph of this solution after the addition of 20.0 ml of 1.00 m naoh solution is 4.71.

Learn more here:

https://brainly.com/question/15578695

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