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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Answer from: 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: