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

Answer:

9.07

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:

$M=\frac{#mol}{L}$

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

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

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:

$M=\frac{0.024~mol}{0.0125~L}=~0.192~M~of~NH_3$

$M=\frac{0.064~mol}{0.0125~L}=~0.512~M~of~NH_4^+$

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