Chemistry, 22.02.2021 15:00 skylarrrrr7663

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 equivalence. The pK, of ammonia is 4.75.
Round your answer to 2 decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added.
pH ?

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Answer from: hailey200127

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

$x \approx 1.19929 \times 10^{-5}$ .

Hence, the concentration of $\rm H^{+}$ at the equivalence point of this titration would be approximately $1.19929 \times 10^{-5}\; \rm M$ .

Hence, the at the equivalence point of this titration would be:

$\begin{aligned}pH &= -\log_{10}[{\rm {H}^{+}}] \\ &\approx -\log_{10} \left(1.19929 \times 10^{-5}\right) \approx 4.92\end{aligned}$ .

Answer from: Quest

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

Answer from: Quest

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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....