Answer: The pH at the equivalence point for the titration will be 0.65.
Solution:
Let the concentration of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
be x
Initial concentration of ![[CH3NH_2]](https://tex.z-dn.net/?f=%5BCH3NH_2%5D)
, c = 0.230 M
at eq'm c-x x x
Expression of 
:
![K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5B%2BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D%3D%5Cfrac%7Bx%5Ctimes%20x%7D%7Bc-x%7D%3D%5Cfrac%7Bx%5E2%7D%7Bc-x%7D)
Since ,methyl-amine is a weak base,c>>x so 
.
Solving for x, we get:
Given, HCl with 0.230 M , it dissociates fully in water which means ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
= 0.230 M
![[OH^-]=[H^+]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BH%5E%2B%5D)
will result in neutral solution, since ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3C%5BH%5E%2B%5D)
Remaining after neutralizing ions
![[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D%5BH%5E%2B%5D-%5BOH%5E-%5D%3D0.230-1.07%5Ctimes%2010%5E%7B-2%7D%3D0.2193%20M)
![pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65](https://tex.z-dn.net/?f=pH%3D-log%7B%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D-log%280.2193%29%3D0.65)
The pH at the equivalence point for the titration will be 0.65.
Compound A : SnCl2
the mass ratio of Cl and Sn: 71 g Cl/ 119 g Sn = 0.579 g Cl/ 1 g Sn
Compound B: the ratio of mass of chlorine per 1 g Sn is twice the ratio of chlorine per 1 g Sn in compound A => 1,194 g Cl/ 1 g Sn = 142,086 g Cl/ 119 g Sn
=> the number of chlorine/ 1 atom Sn = 142.086/ 35.5 = 4.
=> the formula of B: SnCl4
The reaction between KOH and HBr is as follows ;
KOH + HBr ---> H₂O + KBr
Stoichiometry of base to acid is 1:1 molar ratio
Both are strong acid and strong base therefore complete ionization takes place
The number of KOH moles added - 0.50 M / 1000 mL/L x 22 mL = 0.011 mol
the number of HBr moles - 0.25 M /1000 mL/L x 44 mL = 0.011 mol
the number of H⁺ ions and OH⁻ ions are equal therefore the whole amount of acid has been completely neutralised by base.
No remaining acid nor base, therefore solution is neutral.
pH = 7
thats the pH value for a neutral solution
Answer:
Explanation: As already mentioned that the dissolution of ammonium nitrate in water is an endothermic process, which explains that more energy would be needed to break the bond between the reactants rather than the formation of the products.
Hence, the enthalpy of the solution for ammonium nitrate would be positive in nature as energy is being given to the reactants molecules to get converted into the products.
