Calculate the molar solubility of agcl in 0.05m nacl. (ksp = 1.77 x 10-10)

3.54 x 10^{-9}

Explanation:

For this Case we have keep in mind the effect of the Common Ion, which is Cl.

You can find the amount of silver ions at equilibrium from Ksp since You already know the chloride ion concentration. Keep in mind that only a few amount of silver chloride will be dissolved. This means the chloride concentration will stay in 0.05 M, Then I can use Ksp to solve for the silver ion concentration. So you can use the following equation.

AgCl Ksp = Ksp/[Cl^-]= (1.77 x 10^{-10})/(0.05) = 3.54 x 10^{-9}

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

there are two significant figures.