Respuesta :
Answer:
Approximately [tex]4.92[/tex].
Explanation:
Initial volume of the solution: [tex]V = 190.0\; \rm mL = 0.1900\; \rm L[/tex].
Initial quantity of [tex]\rm NH_3[/tex]:
[tex]\begin{aligned} n({\rm NH_3}) &= c({\rm NH_3}) \cdot V({\rm NH_3}) \\ &= 0.3733\; \rm mol \cdot L^{-1} \times 0.1900\; \rm L \\ &\approx 0.154375\; \rm mol\end{aligned}[/tex].
Ammonia [tex]\rm NH_3[/tex] reacts with hydrochloric [tex]\rm HCl[/tex] acid at a one-to-one ratio:
[tex]\rm NH_3 + HCl \to NH_4 Cl[/tex].
Hence, approximately [tex]n({\rm HCl}) = 0.154375\; \rm mol[/tex] of [tex]\rm HCl\![/tex] molecules would be required to exactly react with the [tex]\rm NH_3\![/tex] in the original solution and hence reach the equivalence point of this titration.
Calculate the volume of that [tex]0.3733\; \rm mol \cdot L^{-1}[/tex] [tex]\rm HCl[/tex] solution required for reaching the equivalence point of this titration:
[tex]\begin{aligned}V({\rm HCl}) &= \frac{n({\rm HCl})}{c({\rm HCl})} \\ &\approx \frac{0.154375\; \rm mol}{0.3733\; \rm mol \cdot L^{-1}} \approx 0.413541\; \rm L\end{aligned}[/tex].
Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately [tex]0.413541\; \rm L + 0.1900\; \rm L \approx 0.6035\; \rm L[/tex].
If no hydrolysis took place, [tex]0.154375\; \rm mol[/tex] of [tex]\rm NH_4 Cl[/tex] would be produced. Because [tex]\rm NH_4 Cl\![/tex] is a soluble salt, the solution would contain [tex]0.154375\; \rm mol\![/tex] of [tex]\rm {NH_4}^{+}[/tex] ions. The concentration of [tex]\rm {NH_4}^{+}\![/tex] would be approximately:
[tex]\begin{aligned}c({\rm {NH_4}^{+}}) &= \frac{n({\rm {NH_4}^{+}})}{V({\rm {NH_4}^{+}})}\\ &\approx \frac{0.154375\; \rm mol}{0.6035\; \rm L} \approx 0.255782\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
However, because [tex]\rm NH_3 \cdot H_2O[/tex] is a weak base, its conjugate [tex]\rm {NH_4}^{+}[/tex] would be a weak base.
[tex]\begin{aligned}pK_{\rm a}({{\rm NH_4}}^{+}) &= pK_{\rm w} - pK_{\rm b}({\rm NH_3})\\ &\approx 13.99 - 4.75 = 9.25\end{aligned}[/tex].
Hence, the following reversible reaction would be take place in the solution at the equivalence point:
[tex]\rm {NH_4}^{+} \rightleftharpoons NH_3 + H^{+}[/tex].
Let [tex]x\; \rm mol \cdot L^{-1}[/tex] be the increase in the concentration of [tex]\rm H^{+}[/tex] in this solution because of this reversible reaction. (Notice that [tex]x \ge 0[/tex].) Construct the following [tex]\text{RICE}[/tex] table:
[tex]\begin{array}{c|ccccc} \textbf{R}& \rm {\rm NH_4}^{+} & \rightleftharpoons & {\rm NH_3}& + & {\rm H}^{+}\\ \textbf{I} & 0.255782 \; \rm M \\ \textbf{C} & -x \;\rm M & & + x\;\rm M & & + x\; \rm M \\ \textbf{E} & (0.255782 - x)\; \rm M & & x\; \rm M & & x\; \rm M\end{array}[/tex].
Thus, at equilibrium:
- Concentration of the weak acid: [tex][{\rm {NH_4}^{+}}] \approx (0.255782 - x) \; \rm M[/tex].
- Concentration of the conjugate of the weak acid: [tex][{\rm NH_3}] = x\; \rm M[/tex].
- Concentration of [tex]\rm H^{+}[/tex]: [tex][{\rm {H}^{+}}] \approx x\; \rm M[/tex].
[tex]\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^{+}}]}{[{ \rm {NH_4}^{+}}]} = 10^{pK_\text{a}({\rm {NH_4}^{+}})}[/tex].
[tex]\displaystyle \frac{x^2}{0.255782 - x} \approx 10^{-9.25}[/tex]
Solve for [tex]x[/tex]. (Notice that the value of [tex]x\![/tex] is likely to be much smaller than [tex]0.255782[/tex]. Hence, the denominator on the left-hand side [tex](0.255782 - x) \approx 0.255782[/tex].)
[tex]x \approx 1.19929 \times 10^{-5}[/tex].
Hence, the concentration of [tex]\rm H^{+}[/tex] at the equivalence point of this titration would be approximately [tex]1.19929 \times 10^{-5}\; \rm M[/tex].
Hence, the [tex]pH[/tex] at the equivalence point of this titration would be:
[tex]\begin{aligned}pH &= -\log_{10}[{\rm {H}^{+}}] \\ &\approx -\log_{10} \left(1.19929 \times 10^{-5}\right) \approx 4.92\end{aligned}[/tex].
