Chemistry, 23.07.2020 14:01, phancharamachasm

Suppose you titrate 25.00 mL of 0.200 M KOBr with 0.200M H2SO4. The pH at half-equivalence point is 7.75 a). What is the initial pH of the 25.00mL of 0.200M KOBr mentioned above

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answered: emanihackle9

Approximately 10.88 .

Explanation:

Equilibrium constant

$\rm OBr^{-}$ can act as a weak Bronsted-Lowry base:

$\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)$ .

(Side note: the state symbol of $\rm HOBr$ in this equation is $\rm (l)$ (meaning liquid) because $\rm HOBr$ is a weak acid.)

However, the equilibrium constant of this reaction, $K_\text{eq}$ , isn't directly given. The idea is to find $K_\text{eq}$ using the $\rm pH$ value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same $K_\text{eq}$ value could also be used to find the $\rm pH$ of the solution before the acid was added.

At equilibrium:

$\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}$ .

At the half-equivalence point of this titration, exactly half of the base, $\rm OBr^{-}$ , has been converted to its conjugate acid, $\rm HOBr$ . Therefore, the half-equivalence concentration of $\rm OBr^{-}$ and $\rm HOBr$ should both be equal to one-half the initial concentration of $\rm OBr^{-}$ .

As a result, the half-equivalence concentration of $\rm OBr^{-}$ and $\rm HOBr$ should be the same. The expression for $K_\text{eq}$ can thus be simplified:

$\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}$ .

In other words, the $K_\text{eq}$ of this system is equal to the $\rm OH^{-}$ concentration at the half-equivalence point. Assume that $\rm p\mathnormal{K}_\text{w}$ the self-ionization constant of water, is . The concentration of $\rm OH^{-}$ can be found from the $\rm pH$ value:

$\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}$ .

Therefore, $\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}$ .

Initial pH of the solution

Again, since $\rm KOBr$ is a soluble salt, all that $0.200\; \rm M$ of $\rm KOBr$ in this solution will be in the form of $\rm K^{+}$ and $\rm OBr^{-}$ ions. Before any hydrolysis takes place, the concentration of $\rm OBr^{-}$ should be equal to that of $\rm KOBr$ . Therefore:

$\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M$ .

Let the equilibrium concentration of $[\rm OH^{-}\; (aq)]$ be $x\; \rm M$ . Create a RICE table for this reversible reaction:

$\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}$ .

Assume that external factors (such as temperature) stays the same. The $K_\text{eq}$ found at the half-equivalence point should apply here, as well.

$\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}$ .

At equilibrium:

$\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}$ .

Assume that is much smaller than 0.200 , such that the denominator is approximately the same as 0.200 :

$\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}$ .

That should be equal to the equilibrium constant, $K_\text{eq}$ . In other words:

$\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}$ .

Solve for :

$x \approx 3.35\times 10^{-4}$ .

In other words, the $\rm OH^{-}$ before acid was added was approximately $3.35\times 10^{-4}\; \rm M$ , which is the same as $3.35\times 10^{-4}\; \rm mol \cdot L^{-1}$ . Again, assume that $\rm p\mathnormal{K}_\text{w} = 14$ . Calculate the $\rm pH$ of that solution:

$\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}$ .

(Rounded to two decimal places.)

answered: Guest

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

answered: Guest

it is a good conductor

explanation: