Answer:
Ksp=1.9×10⁻⁸
Explanation:
First write the balane chemical reaction
AgOH + HCl → AgCl+H2O
Titration reaction occur completely
N-factor of AgOH and HCl is one
normality=molarity
Noermality of HCl=0.0136
N₁ and N₂ are the normality of AgOH and HCl respectively
V₁ and V₂ are the volume of AgOH and HCl respectively
mili equvalent of AgOH-mili equivalent of HCl
[tex]N_1[/tex]=?
[tex]V_1[/tex]=250ml
[tex]N_2[/tex]=0.0136
[tex]V_2[/tex]=2.6ml
[tex]N_1V_1=N_2V_2[/tex]
after puttting all the value we get,
[tex]N_1[/tex]=0.000141 =normality of AgOH solution
molarity=0.000141M
Calculation of Ksp
AgOH⇄Ag⁺ + OH⁻
Lets solubility=S
[tex][Ag^+][/tex]=S
[OH⁻]=S
and S is equal to the concentration of AgOH because that concentration is calculated when solution become satrated.
hence S=0.000141M
Ksp=[Ag+][OH-]
Ksp=S×S
Ksp=1.9×10⁻⁸
According to the question,
The chemical reaction is:
As we know the relation,
→ [tex]C_1 V_1 = C_2 V_2[/tex]
(AgOH) (HCl)
By substituting the values, we get
→ [tex]C_1\times 250 \ mL = 0.0136\times 2.60 \ mL[/tex]
[tex]C_1 = \frac{0.0136\times 2.60}{250}[/tex]
[tex]= 0.000141[/tex]
Now,
(s) (s)
hence,
The Ksp will be:
= [tex]s^2[/tex]
= [tex](0.000141)^2[/tex]
= [tex]1.98\times 10^{-8}[/tex]
Thus the solution above is right.
Learn more about titration here:
https://brainly.com/question/16933216
