A 250 mL sample of saturated AgOH solution was titrated with HCl, and the endpoint was reached after 2.60 mL of 0.0136 M of HCl
1 answer:
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
=?
=250ml
=0.0136
=2.6ml
after puttting all the value we get,
=0.000141 =normality of AgOH solution
molarity=0.000141M
<em>Calculation of Ksp</em>
AgOH⇄Ag⁺ + OH⁻
Lets solubility=S
<em /><em>=</em>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
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