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adell [148]2 years ago

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Explanation:

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A chemist titrates 190.0 mL of a 0.8125 M ammonia (NH) solution with 0.3733 M HCl solution at 25 °C. Calculate the pH at equival

stealth61 [152]

Answer:

Approximately $4.92$ .

Explanation:

Initial volume of the solution: $V = 190.0\; \rm mL = 0.1900\; \rm L$ .

Initial quantity of $\rm NH_3$ :

$\begin{aligned} n({\rm NH_3}) &= c({\rm NH_3}) \cdot V({\rm NH_3}) \\ &= 0.3733\; \rm mol \cdot L^{-1} \times 0.1900\; \rm L \\ &\approx 0.154375\; \rm mol\end{aligned}$ .

Ammonia $\rm NH_3$ reacts with hydrochloric $\rm HCl$ acid at a one-to-one ratio:

$\rm NH_3 + HCl \to NH_4 Cl$ .

Hence, approximately $n({\rm HCl}) = 0.154375\; \rm mol$ of $\rm HCl\!$ molecules would be required to exactly react with the $\rm NH_3\!$ in the original solution and hence reach the equivalence point of this titration.

Calculate the volume of that $0.3733\; \rm mol \cdot L^{-1}$ $\rm HCl$ solution required for reaching the equivalence point of this titration:

$\begin{aligned}V({\rm HCl}) &= \frac{n({\rm HCl})}{c({\rm HCl})} \\ &\approx \frac{0.154375\; \rm mol}{0.3733\; \rm mol \cdot L^{-1}} \approx 0.413541\; \rm L\end{aligned}$ .

Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately $0.413541\; \rm L + 0.1900\; \rm L \approx 0.6035\; \rm L$ .

If no hydrolysis took place, $0.154375\; \rm mol$ of $\rm NH_4 Cl$ would be produced. Because $\rm NH_4 Cl\!$ is a soluble salt, the solution would contain $0.154375\; \rm mol\!$ of $\rm {NH_4}^{+}$ ions. The concentration of $\rm {NH_4}^{+}\!$ would be approximately:

$\begin{aligned}c({\rm {NH_4}^{+}}) &= \frac{n({\rm {NH_4}^{+}})}{V({\rm {NH_4}^{+}})}\\ &\approx \frac{0.154375\; \rm mol}{0.6035\; \rm L} \approx 0.255782\; \rm mol \cdot L^{-1}\end{aligned}$ .

However, because $\rm NH_3 \cdot H_2O$ is a weak base, its conjugate $\rm {NH_4}^{+}$ would be a weak base.

$\begin{aligned}pK_{\rm a}({{\rm NH_4}}^{+}) &= pK_{\rm w} - pK_{\rm b}({\rm NH_3})\\ &\approx 13.99 - 4.75 = 9.25\end{aligned}$ .

Hence, the following reversible reaction would be take place in the solution at the equivalence point:

$\rm {NH_4}^{+} \rightleftharpoons NH_3 + H^{+}$ .

Let $x\; \rm mol \cdot L^{-1}$ be the increase in the concentration of $\rm H^{+}$ in this solution because of this reversible reaction. (Notice that $x \ge 0$ .) Construct the following $\text{RICE}$ table:

$\begin{array}{c|ccccc} \textbf{R}& \rm {\rm NH_4}^{+} & \rightleftharpoons & {\rm NH_3}& + & {\rm H}^{+}\\ \textbf{I} & 0.255782 \; \rm M \\ \textbf{C} & -x \;\rm M & & + x\;\rm M & & + x\; \rm M \\ \textbf{E} & (0.255782 - x)\; \rm M & & x\; \rm M & & x\; \rm M\end{array}$ .

Thus, at equilibrium:

Concentration of the weak acid: $[{\rm {NH_4}^{+}}] \approx (0.255782 - x) \; \rm M$ .
Concentration of the conjugate of the weak acid: $[{\rm NH_3}] = x\; \rm M$ .
Concentration of $\rm H^{+}$ : $[{\rm {H}^{+}}] \approx x\; \rm M$ .

$\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^{+}}]}{[{ \rm {NH_4}^{+}}]} = 10^{pK_\text{a}({\rm {NH_4}^{+}})}$ .

$\displaystyle \frac{x^2}{0.255782 - x} \approx 10^{-9.25}$

Solve for $x$ . (Notice that the value of $x\!$ is likely to be much smaller than $0.255782$ . Hence, the denominator on the left-hand side $(0.255782 - x) \approx 0.255782$ .)