Please help!!!!!!!!!!!!
2 answers:
You might be interested in
Answer:
the solubility of CaCO3 is 0.015g/l 25 °C
is favored at equilibrium
Explanation:
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?
solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)
CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)
in partial dissociation , we can say
2.25x 10^-8=
let Ca^2+=CO3^-2=S
2.25x10^-8=S*S
S^2=2.25x10^-8
S=0.00015mol/L
Converting that to g/l
the relative molecular mass of CaCO3=100g/mol
0.00015*100g/mol
0.015g/l
the solubility of CaCO3 is 0.015g/l @room temperature
is favored at equilibrium
<u>Answer:</u> The unknown salt is NaF
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
Moles of salt = 0.050 moles
Volume of solution = 0.500 L
Putting values in above equation, we get:
- To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:
pH + pOH = 14
We are given:
pH = 8.08
- To calculate pOH of the solution, we use the equation:
Putting values in above equation, we get:
The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH
The chemical equation for the hydrolysis of ions follows:
<u>Initial:</u> 0.1
<u>At eqllm:</u> 0.1-x x x
Concentration of
The expression of for above equation follows:
Putting values in above expression, we get:
- To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:
where,
= Ionic product of water =
= Acid dissociation constant
= Base dissociation constant =
Putting values in above equation, we get:
We know that:
So, the calculated is approximately equal to the
of HF
Hence, the unknown salt is NaF