Answer:

Explanation:
Hello!
In this case, since the pH of the given metal is 10.15, we can compute the pOH as shown below:

Now, we compute the concentration of hydroxyl ions in solution:
![[OH^-]=10^{-pOH}=10^{-3.95}=1.41x10^{-4}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-pOH%7D%3D10%5E%7B-3.95%7D%3D1.41x10%5E%7B-4%7DM)
Now, since this hydroxide has the form MOH, we infer the concentration of OH- equals the concentration of M^+ at equilibrium, assuming the following ionization reaction:

Whose equilibrium expression is:
![Ksp=[M^+][OH^-]](https://tex.z-dn.net/?f=Ksp%3D%5BM%5E%2B%5D%5BOH%5E-%5D)
Therefore, the Ksp for the saturated solution turns out:

Best regards!
Answer:
Lead (II) iodide
Explanation:
The reaction of lead (II) nitrate, Pb(NO₃)₂ with KI is:
Pb(NO₃)₂(aq) + 2KI(aq) → KNO₃(aq) + PbI₂(s)
This is a typical double-replacement reaction where anions and cations exchange its couple.
All nitrates are solubles, thus, KNO₃ is not the precipitate.
The only possibility of precipitate is PbI₂,
Lead (II) iodide, a yellow and insoluble solid...
Answer:
sorry i donntttt understand make it clear
Answer:
- <em>The pH of the solution is </em><u><em>7</em></u>
Explanation:
<em>The pH</em> is a measure of the acidity of the solutions. It is defined as the negative logarithm of the molar concentration of hydrogen ions (H⁺).
<em>The hydrogen ion concentration of this solution is 1 × 10⁻⁷ M.</em>
Hence:
- pH = - log (1 × 10⁻⁷) = - (-7) = 7
This pH corresponds to a neutral solution (neither acid nor alkaline).
You should remember this relation bwtween pH and acidity/alkaliinity:
- Low pH (0.0 or close) corresponds to strong acids
- HIgh pH (14.0 or close) corresponds to strong bases
- Acids have pH between 0.0 and 7.0
- Bases have pH between 7.0 and 14.0
Blank 1: nothing (to keep 2 total nitrogen)
blank 2: 3 (to make 6 total hydrogen)
blank 3: 2 (to make 2 total nitrogen and 6 total hydrogen)
hope this helps!! :)