Answer:
![PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)](https://tex.z-dn.net/?f=PO_4%5E%7B3-%7D%28aq%29%2B3Ag%5E%2B%28aq%29%5Crightarrow%20Ag_3PO_4%28s%29)
Explanation:
Hello!
In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:
![Cs_3PO_4(aq)+3AgNO_3(aq)\rightarrow Ag_3PO_4(s)+3CsNO_3(aq)](https://tex.z-dn.net/?f=Cs_3PO_4%28aq%29%2B3AgNO_3%28aq%29%5Crightarrow%20Ag_3PO_4%28s%29%2B3CsNO_3%28aq%29)
Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:
![3Cs^+(aq)+PO_4^{3-}(aq)+3Ag^+(aq)+3NO_3^-(aq)\rightarrow Ag_3PO_4(s)+3Cs^+(aq)+3NO_3^-(aq)](https://tex.z-dn.net/?f=3Cs%5E%2B%28aq%29%2BPO_4%5E%7B3-%7D%28aq%29%2B3Ag%5E%2B%28aq%29%2B3NO_3%5E-%28aq%29%5Crightarrow%20Ag_3PO_4%28s%29%2B3Cs%5E%2B%28aq%29%2B3NO_3%5E-%28aq%29)
In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:
![PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)](https://tex.z-dn.net/?f=PO_4%5E%7B3-%7D%28aq%29%2B3Ag%5E%2B%28aq%29%5Crightarrow%20Ag_3PO_4%28s%29)
Best regards!
![pH = 13.5](https://tex.z-dn.net/?f=%20pH%20%3D%2013.5%20)
Explanation:
Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:
![\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}](https://tex.z-dn.net/?f=%20%5Ctext%7BHAc%7D%20%2B%20%5Ctext%7BOH%7D%5E%7B-%7D%20%5Cto%20%5Ctext%7BAc%7D%5E%7B-%7D%20%2B%20%5Ctext%7BH%7D_2%5Ctext%7BO%7D%20)
The mixture would contain
if
undergoes no hydrolysis; the solution is of volume
after the mixing. The two species would thus be of concentration
and
, respectively.
Construct a RICE table for the hydrolysis of
under a basic aqueous environment (with a negligible hydronium concentration.)
![\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}](https://tex.z-dn.net/?f=%20%5Cbegin%7Barray%7D%7Bcccccccc%7D%20%5Ctext%7BR%7D%20%26%20%5Ctext%7BAc%7D%5E%7B-%7D%28aq%29%20%26%2B%20%26%20%5Ctext%7BH%7D_2%5Ctext%7BO%7D%28aq%29%20%26%20%5Cleftrightharpoons%20%26%20%5Ctext%7BHAc%7D%28aq%29%20%26%20%2B%20%26%20%5Ctext%7BOH%7D%5E%7B-%7D%20%28aq%29%5C%5C%20%5Ctext%7BI%7D%20%26%200.10%20%5C%3B%20%5Ctext%7BM%7D%20%26%20%26%20%26%20%26%20%26%20%260.30%20%5C%3B%20%5Ctext%7BM%7D%5C%5C%20%5Ctext%7BC%7D%20%26%20-x%20%5C%3B%20%5Ctext%7BM%7D%26%20%26%20%26%20%26%20%2Bx%20%5C%3B%20%5Ctext%7BM%7D%26%20%26%20%2Bx%20%5C%3B%20%5Ctext%7BM%7D%20%5C%5C%20%5Ctext%7BE%7D%20%26%20%280.10%20-%20x%29%20%5C%3B%20%5Ctext%7BM%7D%20%26%20%26%20%26%20%26%20x%20%5C%3B%20%5Ctext%7BM%7D%20%26%20%26%20%280.30%20%2Bx%29%20%5C%3B%20%5Ctext%7BM%7D%20%5Cend%7Barray%7D%20)
The question supplied the <em>acid</em> dissociation constant
for acetic acid
; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant
for its conjugate base,
. The following relationship relates the two quantities:
![pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})](https://tex.z-dn.net/?f=%20pK_%7Bb%7D%20%28%5Ctext%7BAc%7D%5E%7B-%7D%29%20%3D%20pK_%7Bw%7D%20-%20pK_%7Ba%7D%28%20%5Ctext%7BHAc%7D%29%20)
... where the water self-ionization constant
under standard conditions. Thus
. By the definition of
:
![[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BHAc%7D%20%28aq%29%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%20%28aq%29%5D%20%2F%20%5B%5Ctext%7BAc%7D%5E%7B-%7D%20%28aq%29%20%5D%20%3D%20K_b%20%3D%20%2010%5E%7B-pK_%7Bb%7D%7D%20)
![x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}](https://tex.z-dn.net/?f=%20x%20%5Ccdot%20%280.3%20%2B%20x%29%20%2F%20%280.1%20-%20x%29%20%3D%2010%5E%7B-9.3%7D%20)
![x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}](https://tex.z-dn.net/?f=%20x%20%3D%201.67%20%5Ctimes%2010%5E%7B-10%7D%20%5C%3B%20%5Ctext%7BM%7D%20%5Capprox%200%20%5C%3B%20%5Ctext%7BM%7D%20)
![[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%200.30%20%2Bx%20%5Capprox%200.30%20%5C%3B%20%5Ctext%7BM%7D%20)
![pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5](https://tex.z-dn.net/?f=%20pH%20%3D%20pK_%7Bw%7D%20-%20pOH%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%7B0.30%7D%20%3D%2013.5%20)
The statement is False.
The molecular (chemical) formulas are same in case of isomers but the arrangement of atoms is different in them. The physical properties are determined by the arrangement of atoms within the molecule.
Since the arrangement of atoms is different in case of isomers, the Physical Properties will also be different.