What is the concentration of silver ions where silver iodide, Agl, is in a solution of hydroiodic

Chemistry

1 answer:

jonny [76]3 years ago

6 0

Answer:

$[Ag^+]=2.82x10^{-4}M$

Explanation:

Hello there!

In this case, for the ionization of silver iodide we have:

$AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-]$

Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:

$[I^-]=[H^+]=10^{-3.55}=2.82x10^{-4}M$

Now, we can set up the equilibrium expression as shown below:

$Ksp=8.51x10^{-17}=(x)(x+2.82x10^{-4})$

Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:

$x=[Ag^+]=2.82x10^{-4}M$

Best regards!

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Mercury(ii) oxide (hgo) decomposes to form mercury (hg) and oxygen (o2). the balanced chemical equation is shown below. 2hgo rig

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Decomposition reaction results in the formation of the products by splitting the reactants. Moles of mercury(ii) oxide needed are 15.63 moles.

<h3>What are moles?</h3>

Moles are the ratio of mass and the molar mass of the compound or the molecule.

Moles of oxygen are calculated as:

$\begin{aligned}\rm n &= \rm \dfrac {mass}{molar\; mass}\\\\&= \dfrac{250}{32}\\\\&= 7.8125 \;\rm moles\end{aligned}$

The balanced chemical reaction can be shown as,

$\rm 2HgO \rightarrow 2Hg + O_{2}$

From the reaction, it can be said that 1 mole of oxygen requires 2 moles of mercury oxide.

Moles of mercury oxide are calculated as:

$\begin{aligned} \rm n HgO& = (7.8125 \;\rm moles \; O_{2}) \times (\dfrac{2 \;\rm moles\; HgO}{1 \;\rm mole \; O_{2}})\\\\\rm n HgO &= 15.625 \;\rm moles \end{aligned}$