Question

A buffer that contains 0.34 M of an acid, HA and 0.37 M of its conjugate base A-, has a pH of 3.26. What is the pH after 0.032 mol of HCl are added to 0.71 L of the solution

Answer 1

Answer: pH ≈ 3.186

Step-by-step explanation:

`p^(H) &=p^(k a)+\left[(A^(-))/(H A)\right] \\3.26 &=p^(k a)+ \log (0.37)/(0.34) \\3.26 &=p^(k a)+\log\ 1.089 \\3.26 &=p^(k e)+0.037 \\p^(k a) &=3.26-0.1105 \\p^(k a) &=3.1495`

`$0.34 \frac{\mathrm{mol}}{2} * 0.71 \mathrm{~L}=0.2414 \ {\mathrm{mol} \rightarrow HA \\\\`

`$0.37 \frac{\mathrm{mol}}{2} * 0.71 \mathrm{~L}=0.2627 \ {\mathrm{mol} \rightarrow A^(-) \\\\`

`$0.2414-0.032 \mathrm{mol}=0.2094 \mathrm{\ mol} \ HA`

`$0.2627+0.032 \mathrm{mol}=0.2947 \mathrm{\ mol} \ A^(-)`

`\begin{aligned}&p H=3.1495+\log(0.2627)/(0.2414) \\&p H=3.1495+0.0367 \\&p H=3.186\end{aligned}`

Therefore, the pH after 0.032 mol of HCl are added to 0.71 L of the solution is approximately 3.186