Chemistry, 22.04.2020 23:15, marmar72

The pH at 25 °C of an aqueous solution of the sodium salt of p-monochlorophenol (NaC6H4ClO) is 11.05. Calculate the concentration of C6H4ClO- in this solution, in moles per liter. Ka for HC6H4ClO is equal to 6.6×10-10.

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answered: dm21930

Approximately $8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}$ .

Explanation:

The K_a in this question refers the dissociation equilibrium of $\rm HC_6H_4ClO$ as an acid:

$\rm HC_6H_4ClO\, (aq) \rightleftharpoons H^{+} \, (aq) + C_6H_4ClO^{-}\, (aq)$ .

$\displaystyle K_a\left(\mathrm{HC_6H_4ClO}\right) = \frac{\left[\mathrm{H^{+}}\right] \cdot \left[\mathrm{C_6H_4ClO^{-}}\right]}{\left[\mathrm{HC_6H_4ClO}\right]}$ .

However, the question also states that the solution here has a $\rm pH$ of 11.05 , which means that this solution is basic. In basic solutions at $\rm 25\;^\circ C$ , the concentration of $\rm H^{+}$ ions is considerably small (typically less than $10^{-7}\;\rm mol \cdot L^{-1}$ .) Therefore, it is likely not very appropriate to use an equilibrium involving the concentration of $\rm H^{+}$ ions.

Here's the workaround: note that $\rm C_6H_4ClO^{-}\, (aq)$ is the conjugate base of the weak acid $\rm HC_6H_4ClO\, (aq)$ . Therefore, when $\rm C_6H_4ClO^{-}\, (aq)$ dissociates in water as a base, its K_b would be equal to $\displaystyle \frac{K_w}{K_a} \approx \frac{10^{-14}}{K_a}$ . ( K_w is the self-ionization constant of water. $K_w \approx 10^{-14}$ at $\rm 25\;^\circ C$ .)

In other words,

$\begin{aligned} & K_b\left(\mathrm{C_6H_4ClO^{-}}\right) \\ &= \frac{K_w}{K_a\left(\mathrm{HC_6H_4ClO}\right)} \\ &\approx \frac{10^{-14}}{6.6 \times 10^{-10}} \\ & \approx 1.51515 \times 10^{-5}\end{aligned}$ .

And that K_b value corresponds to the equilibrium:

$\rm C_6H_4ClO^{-}\, (aq) + H_2O\, (l) \rightleftharpoons HC_6H_4ClO\, (aq) + OH^{-}\, (aq)$ .

$\displaystyle K_b\left(\mathrm{C_6H_4ClO^{-}}\right) = \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{\left[\mathrm{C_6H_4ClO^{-}}\right]}$ .

The value of K_b has already been found.

The $\rm OH^{-}$ concentration of this solution can be found from its $\rm pH$ value:

$\begin{aligned}& \left[\mathrm{OH^{-}}\right] \\ &= \frac{K_w}{\left[\mathrm{H}^{+}\right]} \\ & = \frac{K_w}{10^{-\mathrm{pH}}} \\ &\approx \frac{10^{-14}}{10^{-11.05}} \\ &\approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1} \end{aligned}$ .

To determine the concentration of $\left[\mathrm{HC_6H_4ClO}\right]$ , consider the following table:

$\begin{array}{cccccc}\textbf{R} &\rm C_6H_4ClO^{-}\, (aq) & \rm + H_2O\, (l) \rightleftharpoons & \rm HC_6H_4ClO\, (aq) & + & \rm OH^{-}\, (aq) \\ \textbf{I} & (?) & \\ \textbf{C} & -x & & + x& & +x \\ \textbf{E} & (?) - x & & x & & x\end{array}$

Before hydrolysis, the concentration of both $\mathrm{HC_6H_4ClO}$ and $\rm OH^{-}$ are approximately zero. Refer to the chemical equation. The coefficient of $\mathrm{HC_6H_4ClO}$ and $\mathrm{HC_6H_4ClO}$ are the same. As a result, this equilibrium will produce $\rm OH^{-}$ and $\mathrm{HC_6H_4ClO}$ at the exact same rate. Therefore, at equilibrium, $\left[\mathrm{HC_6H_4ClO}\right] \approx \left[\mathrm{OH^{-}}\right] \approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1}$ .

Calculate the equilibrium concentration of $\left[\mathrm{C_6H_4ClO^{-}}\right]$ from $K_b\left(\mathrm{C_6H_4ClO^{-}}\right)$ :

$\begin{aligned} & \left[\mathrm{C_6H_4ClO^{-}}\right] \\ &= \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{K_b}\\&\approx \frac{\left(1.1220 \times 10^{-3}\right) \times \left(1.1220 \times 10^{-3}\right)}{1.51515\times 10^{-5}}\; \rm mol \cdot L^{-1} \\ &\approx 8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}\end{aligned}$ .

answered: Guest

1.consider the following balanced equation and answer parts a through. show calculations for each part.

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.

explanation:

answered: Guest

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year