Answer:
9.07
Explanation:
We have to start with the buffer system reaction, so:
[tex]NH_4^+~<->~NH_3~+~H^+[/tex]
When we have the hydrochloric acid (a strong acid) the [tex]H^+[/tex] of the hydrochloric acid ([tex]HCl[/tex]) will interact with the base of the buffer system ([tex]NH_3[/tex]) to produce more acid ([tex]NH_4^+[/tex]), so:
[tex]HCl~+~NH_3->~NH_4^+~+Cl^-[/tex]
Therefore the concentration of [tex]NH_3[/tex] will decrease and the concentration of [tex]NH_4[/tex] will increase. The next step then would be the calculation of the moles of the acid and base in the buffer system. So:
[tex]M=\frac{#mol}{L}[/tex]
[tex]#mol=0.353*0.125=0.044~mol~of~NH_4^+[/tex]
[tex]#mol=0.352*0.125=0.044~mol~of~NH_3[/tex]
If we add 0.02 mol of [tex]HCl[/tex] we can calculate the amount of acid and base that changes in the buffer system.
[tex]0.044~mol~of~NH_3~-~0.02=0.024[/tex]
[tex]0.044~mol~of~NH_4^+~+~0.02=0.064[/tex]
Now, we can calculate the concentration of each species if we divide by the number of moles:
[tex]M=\frac{0.024~mol}{0.0125~L}=~0.192~M~of~NH_3[/tex]
[tex]M=\frac{0.064~mol}{0.0125~L}=~0.512~M~of~NH_4^+[/tex]
If we use the hendersson hasselbach equation we can calculate the pH value again:
[tex]pH=p{ K }_{ a }+log(\frac { { [A }^{ - }] }{ [HA] } )[/tex]
[tex]pH=9.5+log(\frac {0.192}{0.512} )[/tex]
[tex]pH=9.07[/tex]
The final pH value is 9.07
