Question

A buffer solution contains 0.353 M ammonium bromide and 0.352 M ammonia. If 0.0200 moles of hydrochloric acid are added to 125 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume change does not change upon adding hydrochloric acid)

Answer 1

9.07

Step-by-step explanation:

We have to start with the buffer system reaction, so:

`NH_4^+~<->~NH_3~+~H^+`

When we have the hydrochloric acid (a strong acid) the
`H^+` of the hydrochloric acid (
`HCl`) will interact with the base of the buffer system (
`NH_3`) to produce more acid (
`NH_4^+`), so:

`HCl~+~NH_3->~NH_4^+~+Cl^-`

Therefore the concentration of
`NH_3` will decrease and the concentration of
`NH_4` will increase. The next step then would be the calculation of the moles of the acid and base in the buffer system. So:

`M=(#mol)/(L)`

`#mol=0.353*0.125=0.044~mol~of~NH_4^+`

`#mol=0.352*0.125=0.044~mol~of~NH_3`

If we add 0.02 mol of
`HCl` we can calculate the amount of acid and base that changes in the buffer system.

`0.044~mol~of~NH_3~-~0.02=0.024`

`0.044~mol~of~NH_4^+~+~0.02=0.064`

Now, we can calculate the concentration of each species if we divide by the number of moles:

`M=(0.024~mol)/(0.0125~L)=~0.192~M~of~NH_3`

`M=(0.064~mol)/(0.0125~L)=~0.512~M~of~NH_4^+`

If we use the hendersson hasselbach equation we can calculate the pH value again:

`pH=p{ K }_( a )+log(\frac { { [A }^( - )] }{ [HA] } )`

`pH=9.5+log(\frac {0.192}{0.512} )`

`pH=9.07`

The final pH value is 9.07